A_{∞}-categories

Matthias Künzer Universität Stuttgart

January 24, 2017

Inhalt

1 Kadeishvili  10
1.1 Gradings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.1 Grading categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.2 Graded modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.1.3 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.2 A_{∞}-algebras and A_{∞}-categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.3 The regular differential graded category for complexes . . . . . . . . . . . . . . . . . . . . 26
1.4 Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.5 Getting rid of signs by conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.6 A tensor coalgebra interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
1.7 Kadeishvili’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
2 Schmid’s extension of Kadeishvili  62
2.1 Split-filtered Z-graded modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
2.2 eA_{∞}-algebras and eA_{∞}-categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
2.3 A base of an induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
A Problems and solutions  71
A.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
A.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Introduction

Problem

Suppose given a commutative ring R. Suppose given a finite group G. Suppose given RG-modules X, Y, Z. What RG-modules M have a filtration with subfactors X, Y and Z? I.e. we ask for RG-modules M having a chain M = M_{0} ⊇ M_{1} ⊇ M_{2} ⊇ M_{3} = 0 of submodules such that M_{0}/M_{1} ≃ X and M_{1}/M_{2} ≃ Y and M_{2}/M_{3} ≃ Z.

M = M_{0} M_{1}_{X} M_{2}_{Y} 0 = M_{3}_{Z}

The short exact sequences of the form Z → M_{1} → Y are controlled by Ext_{RG}^{1}(Y, Z). The short exact sequences of the form M_{1} → M → X are controlled by Ext_{RG}^{1}(X, M_{1}). The latter R-module might be difficult to cope with, because we need to use M_{1} as input. It would be preferable to make do with Ext_{RG}^{1}(X, X), Ext_{RG}^{1}(X, Y), Ext_{RG}^{1}(Y, X), …, Ext_{RG}^{1}(Z, Z). The Yoneda product maps e.g.

m_{2} : Ext_{RG}^{1}(X, Y) × Ext_{RG}^{1}(Y, Z) \overline{}Ext_{RG}^{1}(X, Z) .

But to reconstruct modules such as M, we also need higher multiplication maps such as e.g.

m_{3} : Ext_{RG}^{1}(Z, X) × Ext_{RG}^{1}(X, X) × Ext_{RG}^{1}(X, Y) \overline{}Ext_{RG}^{1}(Z, Y) ,

and similarly m_{4} , m_{5} , … These data form an A_{∞}-category. The aimed-for reconstruction will be achieved with Keller’s filt-construction.

Kadeishvili

A cohomology algebra

Given a finite group G, we can consider its cohomology algebra H(G; R) with coefficients in a commutative ring R. (^{1})

This cohomology algebra can be calculated as follows. Let P be a projective resolution of R over RG. Let P^{[k]} arise from P by a shift of k steps to the left, where k ∈ Z_{⩾0} , and by multiplying each differential by (−1)^{k}. Let Ṗ denote the graded RG-module underlying P, forgetting the differentials. Form the graded algebra DG(G; R) having

DG^{k}(G; R) := _{RG-grad}(Ṗ , Ṗ^{[k]}) ,

where the latter stands for the R-module of morphisms in the category RG-grad of graded RG-modules.

Remembering the differentials of P again, DG(G; R) becomes a differential graded algebra. Its cohomology algebra is H(DG(G; R)) = H(G; R).

If R is a field: A quasiisomorphism of A_{∞}-algebras

Consider the case that R is a field. Suppose given a differential graded algebra D over the field R, with differential d = m_{1}^{D} : D = D^{⊗1} \overline{}D and multiplication m_{2}^{D} : D^{⊗2} \overline{}D.

Then its cohomology algebra H(D) carries not only a multiplication map

m_{2}^{H(D)} : H(D)^{⊗2} \overline{}H(D) ,

but also higher multiplication maps

m_{n}^{H(D)} : H(D)^{⊗n} \overline{}H(D) for n ⩾ 3

and

m_{1}^{H(D)} := 0 : H(D)^{⊗1} \overline{}H(D) ,

fitting together to turn H(D) into an A_{∞}-algebra. An A_{∞}-algebra with m_{1} = 0 is called minimal.

But also D can be viewed as a A_{∞}-algebra by letting m_{n}^{D} := 0 : D^{⊗n} \overline{}D for n ⩾ 3.

Kadeishvili’s Theorem states that there is a quasiisomorphism from D to H(D), i.e. a morphism of A_{∞}-algebras

D \overline{}H(D)

that induces an isomorphism on cohomology [1, Th. 1]. More precisely, it states that the A_{∞}-structure on H(D) can be chosen in such a way that such a quasiisomorphism emerges. The resulting A_{∞}-algebra is, of course, determined uniquely up to quasiisomorphism.

1In the literature, H(G; R) is often written H^{∗}(G; R).

The assumption of R being a field is used in this process to ensure that every surjective R-linear map is a retraction. This prevents us from directly generalising to R being a discrete valuation ring, say.

In particular, Kadeishvili’s Theorem can be applied to D = DG(G; R) and H(D) = H(G; R) in the case of R = F_{p}, where p is a prime divisor of |G|, but not in the case of R = Z_{(p)} .

Generalisation to arbitrary ground rings

To generalise to an arbitrary commutative ring R, we replace the cohomology modules by projective resolutions over R. I.e. given a differential graded algebra D, we choose an augmented projective resolution

… \overline{}P_{2}^{i} \overline{}P_{1}^{i} \overline{}P_{0}^{i} \overline{}H^{i}(D) \overline{}0 ,

as suggested by Keller.

Schmid’s Theorem states, roughly put, that there exists a minimal eA_{∞}-algebra structure on ⨁_{i,j} P_{j}^{i} such that there exists a quasiisomorphism to D [5, Th. 90]. Here, on the one hand, an eA_{∞}-algebra structure is a refinement of an A_{∞}-algebra structure; on the other hand, eA_{∞}-minimality is a weakening of A_{∞}-minimality.

In particular, Schmid’s Theorem can be applied to D = DG(G; R) and H(D) = H(G; R) in the case of R = Z_{(p)} .

Schmid’s procedure is similar to that of Sagave [3], one of the differences being that Sagave resolves once more in the process, while Schmid sticks to the initially chosen projective resolutions; cf. [3, Th. 1.1, Rem. 4.14].

Modules

From A_{∞}-algebras to A_{∞}-categories

To fix ideas, we consider RG-modules again.

Note that H(G; R) = Ext_{RG}(R, R), where R is the trivial RG-module.

Suppose given RG-modules S_{1} , …, S_{n}. To take these several objects into account, we refine the notion of an A_{∞}-algebra to that of an A_{∞}-category, in that we endow an A_{∞}-algebra with a categorical grading, which is, in a sense, a fixed Peirce decomposition.

If R is a field, the categorical version of Kadeishvili’s Theorem establishes the structure of a minimal A_{∞}-category on ⨁_{α, β ∈ [1,n]} Ext_{RG}(S_{α}, S_{β}).

Over arbitrary R, the categorical version of Schmid’s Theorem establishes the structure of a minimal eA_{∞}-category on ⨁_{α, β ∈ [1,n]} P_{α,β}, where P_{α,β} is a projective resolution of Ext_{RG}(S_{α}, S_{β}) over R.

The filt-construction

A finitely generated RG-module is called (S_{α})_{α}-filtered, if it has a filtration whose subfactors occur in (S_{α})_{α}, up to isomorphy, repetition allowed, ordered arbitrarily.

If R is a field, Keller’s filt-construction recovers the full subcategory of (S_{α})_{α}-filtered modules in RG -mod in terms of the A_{∞}-category ⨁_{α, β ∈ [1,n]} Ext_{RG}(S_{α}, S_{β}) ; cf. [2, §7.7, Theorem]. In particular, if the modules S_{α} represent the isoclasses of simple modules, we recover the whole category RG -mod.

Schmid generalised this to arbitrary R, using the eA_{∞}-category ⨁_{α, β ∈ [1,n]} P_{α,β}; cf. [5, Th. 131]. (^{2})

A desirable future application

We can e.g. take G = S_{n} and R = Z_{(p)} for some prime divisor of n! and let S_{α} run through Specht modules, or certain submodules thereof. One might ask whether, in small examples, the shape of the indecomposable projective modules can be explained through the said eA_{∞}-category; just as for Z_{(5)}S_{5} , the shape of certain indecomposable projective modules can be explained as being glued from two Specht modules via an element of Ext^{1} ; cf. [4, Ex. 7].

Organisatorial matters

We essentially follow the master thesis of Stephan Schmid [5]. Responsibility for mistakes and obscurities in this script remains with me. I will be thankful for any hints on this matter.

We presuppose Algebra and some basic knowledge from Homological Algebra, which will be recalled in the exercises if necessary.

Sometimes we refer to exercises and solutions, so they are to be viewed as part of the script.

Because of running modifications, it is recommended to work with the file during the semester and to print a paper copy only afterwards.

Stuttgart, winter semester 2016/17

Matthias Künzer

2Actually, conditions on R do not play a role in this assertion on the equivalence from the filtconstruction to the category of filtered modules; Schmid gives a somewhat straightened proof and ensures that the equivalence from the filt-construction to filtered modules can be applied to his eA_{∞}-category ⨁_{α, β ∈ [1,n]} P_{α,β} .

Conventions

Let C be a category.

• Given a, b ∈ Z, we write

[a, b] := {z ∈ Z : a ⩽ z ⩽ b}

for the integral interval.

• We stipulate that −∞ < a < ∞ for a ∈ Z. We write

\begin{aligned}[a, ∞] & := & \{z ∈ Z : a ⩽ z\} ∪ \{∞\} \\ [−∞, a] & := & \{−∞\} ∪ \{z ∈ Z : z ⩽ a\} . \end{aligned}

• XXX Z_{⩾n}, Z_{<n}

• Given a set X, “for x ∈ X ” means “for all x ∈ X ”.

• We use the symbol ⊔ for the (interior and exterior) disjoint union of sets and for the concatenation of tuples.

• All categories are suppose to be small (with respect to a given universe). I.e. we have the sets Mor(C) and Ob(C).

• We have source and target maps, s_{C} , t_{C} : Mor(C) → Ob(C), respectively, mapping a morphism X ⟶^{f} Y in C to fs_{C} = X and to ft_{C} = Y, respectively.

• For X ∈ Ob(C), we write id = id_{X} for the identity morphism on X. In some contexts, we also write 1 = 1_{X} = id_{X}.

• Given a category C and X, Y ∈ Ob(C), we write _{C}(X, Y) = { f ∈ Mor(C) : fs_{C} = X and ft_{C} = Y }.

• Given k ⩾ 0, we write X^{×k} := ∏_{i∈[1,k]} X for the k-fold cartesian product. We identify along X^{×1} → X, (x) ↦ x. Moreover, X^{×0} = { () } is a one-element set.

• XXX tensor product ⨂_{i∈I} M_{i}, elementary tensors (m_{i})_{i∈I}^{⊗}; if I = [a, b] interval, then ⨂_{i∈[a,b]} M_{i} =: M_{a} ⊗ M_{a+1} ⊗ ⋯ ⊗ M_{b} and (m_{i})_{i∈I}^{⊗} =: m_{a} ⊗ m_{a+1} ⊗ ⋯ ⊗ m_{b} XXX as far as possible just as for cartesian products

• XXX tensor product of R-modules associative via identification, additive via identification, R ⊗ M = M via identification

• XXX concerning tensor products, we freely use [Ritter]

• XXX abbreviate v_{1} ⊗ … ⊗ v_{k} = v_{[1,k]}^{⊗} XXX • XXXR-linear preadditive categoryXXX plus Example: Peirce decomposition [Rit-

ter]

• XXX composition of morphisms naturally, composition of functors traditionally, with some exceptions, e.g. for certain standard maps, for maps written in index notation or for shift functors z ↦ z[i] XXX

• XXX Let Cat be the (1-)category of categories. Let Set be the category of set

• XXX poset: partially ordered set, category of posets and monotone maps: Poset, (full) subposet (full may be omitted), X_{⩽ξ}

• XXX category of functors C, D .

• XXX terminal category !

• XXX inverse often f^{−}

• XXX automorphism of a category

• XXX complex

• XXX exact, short exact in B -Mod

• XXX augmented projective resolution

Fixing the ground ring R

Let R be a commutative ring.

By a module we understand an R-module.

By a linear map we understand an R-linear map.

We write ⊗ := ⊗_{R} .

By an algebra we understand an R-algebra.

By a linear category we understand an R-linear preadditive category.

By a linear additive category we understand an R-linear additive category.

By a linear functor we understand an R-linear additive functor.

Chapter 1

Kadeishvili

1.1 Gradings

1.1.1 Grading categories

Let Z be a category.

Example 1

(1) Let G be a group. Then, by abuse of notation, G can be considered as a category with Ob(G) = {G}, Mor(G) = G and composition given by multiplication.

(2) We may specialise (1) to G = Z. So Ob(Z) = {Z} and Mor(Z) = Z, composition being given by addition. E.g. we get the commutative triangle

\xymatrix{Z \ar[rr]^{8} \ar[dr]_{3} &  & Z \\  & Z \ar@{-->}[ur]_{5}}

(3) Conversely, if | Ob(Z)| = 1 and each morphism in Z is an isomorphism, we may consider Z as a group. More precisely, Mor(Z) is a group with multiplication given by composition.

More generally, if | Ob(Z)| = 1, we may consider Z to be a monoid.

(4) Let I be a set. By abuse of notation, let I^{×2} denote the pair category on I, having Ob(I^{×2}) = I and Mor(I^{×2}) = I^{×2}.

A morphism (i, j) ∈ Mor(X^{×2}) has source (i, j)s_{I^{×2}} = i and target (i, j)t_{I^{×2}} = j.

The composite of the morphisms

x ⟶^{(x,y)} y ⟶^{(y,z)} z

is

x ⟶^{(x,z)} z .

So the identity on x ∈ Ob(X^{×2}) is id_{x} = (x, x).

(5) Write Ob^{×2}(Z) := (Ob(Z))^{×2} for the pair category on the set Ob(Z). Definition 2 A shift S on Z is a tuple of maps

S = (_{Z}(X, Y) ⟶^{S_{X,Y}} _{Z}(X, Y))_{X, Y ∈ Ob(Z)}

such that properties (1, 2) hold.

(1) The map S_{X,Y} is bijective for X, Y ∈ Ob(Z).

(2) Given X ⟶^{a} Y ⟶^{b} Z in Z, we have (a · b)S_{X,Z} = aS_{X,Y} · b = a · bS_{Y,Z} .

We often write aS := aS_{X,Y} for X ⟶^{a} Y in Z. We often write a[k] := aS^{k} for X ⟶^{a} Y in Z and k ∈ Z.

Note that S is not required to be a functor.

Example 3 We have the identical shift (id_{_{Z}(X,Y)})_{X, Y ∈ Ob(Z)} on Z. Definition 4 Suppose given a shift S on Z. A degree function on (Z, S) is a map deg : Mor(Z) → Z such that properties (1, 2) hold.

(1) Given X ⟶^{a} Y ⟶^{b} Z in Z, we have (a · b) deg = a deg +b deg .

(2) Given X ⟶^{a} Y in Z, we have

(aS) deg = a deg +1 .

So (a[k]) deg = a deg +k for X ⟶^{a} Y in Z and k ∈ Z. Definition 5 The category Z, together with a shift S on Z and a degree function deg on (Z, S), is called a grading category. Cf. Definitions 2 and 4. We often abbreviate Z = (Z, S, deg).

Example 6

(1) The category Z := Z carries the shift (Z ⟶^{i} Z)S := (Z ⟶^{i+1} Z) for i ∈ Z, whence i[k] = i + k for i, k ∈ Z. Then (Z, S) carries the degree function deg = id_{Z} . So i deg = i for i ∈ Z.

(2) We generalise (1). Let C be a category. Consider the category Z := Z × C. The example in (1) can be considered as the particular case C = ! . The category Z carries the shift

(Z ⟶^{i} Z, X ⟶^{a} Y)S := (Z ⟶^{i+1} Z, X ⟶^{a} Y)

for (i, a) = (Z ⟶^{i} Z, X ⟶^{a} Y) in Mor(Z) = Mor(Z) × Mor(C). So (i, a)[k] = (i + k, a) for i, k ∈ Z. For (i, a), (j, b) ∈ Mor(Z) with X ⟶^{a} Y ⟶^{b} Z in C, we obtain in fact the following.

\begin{aligned}((i, a) · (j, b))S & = & (i + j, a · b)S & = & (i + j + 1, a · b) \\ (i, a)S · (j, b) & = & (i + 1, a) · (j, b) & = & (i + j + 1, a · b) \\ (i, a) · (j, b)S & = & (i, a) · (j + 1, b) & = & (i + j + 1, a · b) \end{aligned}

Then (Z, S) carries the degree function

Mor(Z) ^{\overline{}^{deg}} Z
(i, a) \overline{}(i, a) deg := i .

For (i, a), (j, b) ∈ Z × C with X ⟶^{a} Y ⟶^{b} Z in C, we obtain in fact the following.

((i, a) · (j, b)) deg = (i + j, a · b) deg = i + j = (i, a) deg +(j, b) deg
((i, a)S) deg = (i + 1, a) deg = i + 1 = (i, a) deg +1

Definition 7 Suppose given n ∈ Z_{⩾1} . A tuple (y_{i})_{i∈[1,n]} ∈ Mor(Z)^{×n} is called composable if y_{i}t_{Z} = y_{i+1}s_{Z} for i ∈ [1, n − 1]. We often abbreviate \underline{y} = (y_{i})_{i∈[1,n]} .

Definition 8 Suppose given z ∈ Mor(Z) and n ∈ Z_{⩾1} . Let

fact_{n}(z) := { (y_{i})_{i∈[1,n]} ∈ Mor(Z)^{×n} : (y_{i})_{i∈[1,n]} composable and z = y_{1} · y_{2} ⋯ y_{n} }

be the set of factorisations of z of length n.

Example 9 (1) Let z ∈ Mor(Z). Then fact_{3}(z) consists of the diagrams ⟶^{y_{1}} ⟶^{y_{2}} ⟶^{y_{3}} with y_{1}·y_{2}·y_{3} = a.

[figure: z; y_{1}; y_{2}; y_{3}]

(2) Let z ∈ Mor(Z). We have fact_{1}(z) = {z}.

(3) For z ∈ Z, we have fact_{2}(z) = { (y_{1}, y_{2}) ∈ Z^{×2} : y_{1} + y_{2} = z }.

1.1.2 Graded modules

Let Z = (Z, S, deg) be a grading category; cf. Definition 5. Definition 10

(1) A Z-graded module is a map M : Mor(Z) → Ob(R -Mod), z ↦ M^{z}, often written (M^{z})_{z∈Mor(Z)} or just (M^{z})_{z}.

(2) Suppose given a Z-graded module M. Suppose given z ∈ Mor(Z) and m ∈ M^{z}.

We write

m deg := z deg

for the degree of m.

(3) Suppose given Z-graded modules L and M. A (Z-)graded linear map f from L to M is a tuple of linear maps (L^{z} ⟶^{f^{z}} M^{z})_{z∈Mor Z} , often written just (L^{z} ⟶^{f^{z}} M^{z})_{z} or (f^{z})_{z}. So

(L ⟶^{f} M) = (L^{z} ⟶^{f^{z}} M^{z})_{z} .

(4) The category

Z-grad_{0}

has the Z-graded modules as objects and the graded linear maps as morphisms. The composite of the Z-graded linear maps L ⟶^{f} M ⟶^{g} N is given by

f · g = ((f · g)^{z})_{z∈Mor(Z)} := (f^{z} · g^{z})_{z∈Mor(Z)} .

We have the identity id_{M} := (id_{M^{z}})_{z∈Mor(Z)} .

(5) Suppose given a set I and Z-graded modules M_{i} for i ∈ I. Define the (external)

direct sum of the tuple (M_{i})_{i∈I} as the Z-graded module

⨁_{i∈I} M_{i} := ( ⨁_{i∈I} M_{i}^{z})_{z∈Mor(Z)} .

for z ∈ Mor(Z).

Example 11 Let B be a linear category. Let Z := Z × Ob^{×2}(B). Then Mor(Z) = Z × Mor(Ob^{×2}(B)) and so

(\{ \begin{matrix}_{B}(X, Y) & if i = 0 \\ 0 & if i ≠ 0 \end{matrix} \})_{(i,(X,Y)) ∈ Z×Mor(Ob^{×2}(B))}

is a (Z × Ob^{×2}(B))-graded module. In particular, given an algebra B, we obtain a linear category, abusively again denoted by B, having Ob(B) = {B} and Mor(B) = B, composition given by multiplication. Then

(\{ \begin{matrix}B & if i = 0 \\ 0 & if i ≠ 0 \end{matrix} \})_{i ∈ Z}

is a Z-graded module, concentrated in degree 0.

Definition 12 Given a Z-graded module M, we let SM be the Z-graded module defined by

(SM)^{z} := M^{zS}

for z ∈ Mor(Z). Given a Z-graded linear map L ⟶^{f} M, we let SL ⟶^{Sf} SM be the Z-graded linear map defined by

((SL)^{z} ⟶^{(Sf)^{z}} (SM)^{z}) := (L^{zS} ⟶^{f^{zS}} M^{zS})

for z ∈ Mor(Z). We have a functor

Z-grad_{0} ⟶^{S} Z-grad_{0}
(L ⟶^{f} M) ↦ (SL ⟶^{Sf} SM)

This functor is an automorphism of Z-grad_{0} ; cf. Problem4.(1). We often write (L^{[k]} ⟶^{f^{[k]}} M^{[k]}) := (S^{k}L ⟶^{S^{k}f} S^{k}M) for L ⟶^{f} M in Z-grad_{0} and k ∈ Z. So (M^{[k]})^{z} = M^{z[k]} and (f^{[k]})^{z} = f^{z[k]} for z ∈ Mor(Z).

Definition 13 Suppose given Z-graded modules L, M and N. A shift-graded linear map of degree k from L to M is a pair (f, k), where f : L → M^{[k]} is a graded linear map.

So f = (L^{z} ⟶^{f^{z}} M^{z[k]})_{z∈Mor(Z)} .

Write (f, k) deg := k. Suppose given shift-graded linear maps L ⟶^{(f,k)} M ⟶^{(g,ℓ)} N. Then L ⟶^{f} M[k] and M[k] ⟶^{g[k]} N[k+ℓ] in Z-grad_{0} , i.e. as graded linear maps. The composite of (f, k) and (g, ℓ) is defined by

(f, k) · (g, ℓ) := (f · g[k], k + ℓ) : L \overline{}N[k + ℓ] .

We have the identity (id_{M} , 0) on M.

We call (f, k) piecewise injective if f^{z} is injective for z ∈ Mor(Z).

We call (f, k) piecewise surjective if f^{z} is surjective for z ∈ Mor(Z).

The category

Z-grad

has the Z-graded modules as objects and the shift-graded linear maps as morphisms. Cf. Problem 4.(2).

By abuse of notation, we let

(SL ⟶^{S(f,k)} SM) := (SL ⟶^{(Sf,k)} SM)(Sf, k)

for L ⟶^{(f,k)} M in Z-grad. Then S is an automorphism on Z-grad ; cf. Problem 4.(3). Accordingly, we write

(L^{[t]} ⟶^{(f,k)^{[t]}} M^{[t]}) := (L^{[t]} ⟶^{(f^{[t]},k)} M^{[t]})

for t ∈ Z.

Finally, given morphisms (f, k), (g, k) ∈ Mor(Z-grad) of the same degree and r, s ∈ R, we let

r(f, k) + s(g, k) := (rf + sg, k) ;

cf. Problem 4.(4).

Definition.

(1) Suppose given a Z-graded module M.

Suppose given a submodule M̃^{z} ⊆ M^{z} for each z ∈ Mor(Z). Then M̃ := (M̃^{z})_{z∈Mor(Z)} is called a Z-graded submodule of M. We write M̃ ⊆ M. We have the shift-graded linear inclusion map of degree 0

M̃ → M
at z ∈ Mor(Z) : M̃^{z} → M^{z}
m̃ ↦ m̃ .

We may form the factor module M^{z}/M̃^{z} for each z ∈ Mor(Z). Then M/M̃ := (M^{z}/M̃^{z})_{z∈Mor(Z)} is called the Z-graded factor module of M modulo M̃. We have the shift-graded linear residue-class map of degree 0

M → M/M̃
at z ∈ Mor(Z) : M^{z} → M^{z}/M̃^{z}
m ↦ m + M̃^{z} .

(2) Suppose given Z-graded modules L and M. Suppose given a shift-graded linear map L ⟶^{f} M of degree d. Let Kern(f) := (Kern(f^{z}))_{z∈Mor(Z)} . Then Kern(f) is a Z-graded submodule of L. Let Im(f) := (Im(f^{z}))_{z∈Mor(Z)} . Then Im(f) is a Z-graded submodule of M. Let Cokern(f) := (M^{z}/ Im(f^{z}))_{z∈Mor(Z)} . Then Cokern(f) = M/ Im(f). Suppose given Z-graded submodules L̃ ⊆ L and M̃ ⊆ M. Write the inclusions L̃ ⟶^{i} L and M̃ ⟶^{j} M. Suppose that Im(i · f) ⊆ M̃.

There exists a unique shift graded linear map L̃ ⟶^{f|M̃ ; L̃} M̃, called the restriction of f to L̃ in the source and M̃ in the target, making the diagram

\xymatrix{L \ar[r]^{f} & M \\ L̃ \ar[r]^{f|M̃ ; L̃} \ar[u]^{i} & M̃ \ar[u]_{j}}

commutative. In particular, f|M̃ ; L̃ is of degree d. We also write L̃f := Im(f|L̃) ⊆ M̃. If L̃ = L, we also write f|M̃ := f|M̃ ; L. If M̃ = M, we also write f|L̃ := f|M ; L̃.

(3) Suppose given a Z-graded module M. Suppose given a set I and Z-graded submodules M_{i} ⊆ M for i ∈ I. Write +_{i∈I} M_{i} := ( +_{i∈I} M_{i}^{z})_{z} for the (inner) sum of the tuple (M_{i})_{i∈I} of submodules, which is a graded submodule of M. Consider the following shift-graded linear map of degree 0.

⨁_{i∈I} M_{i} ⟶^{φ_{(M_{i})_{i∈I}}}  +_{i∈I} M_{i}
at z ∈ Mor(Z) : ⨁_{i∈I} M_{i}^{z}  →  +_{i∈I} M_{i}^{z}
(m_{i})_{i∈I}  ↦  ∑_{i∈I} m_{i}

We say that +_{i∈I} M_{i} is a(n) (inner) direct sum of (M_{i})_{i∈I} if φ_{(M_{i})_{i∈I}} is an isomor-

phism. In this case, we also write, by abuse of notation, ⨁_{i∈I} M_{i} := +_{i∈I} M_{i}.

So the sum +_{i∈I} M_{i} is direct if and only if the sum +_{i∈I} M_{i}^{z} is direct for z ∈ Mor(Z).

(4) Suppose given a Z-graded module M. Suppose given a set I and Z-graded submodules M_{i} ⊆ M for i ∈ I. Write ⋂_{i∈I} M_{i} := ( ⋂_{i∈I} M_{i}^{z} )_{z}, which is a graded submodule of M.

Remark 14 Let k ∈ Z. Given (f, k) ∈ Mor(Z-grad), we often write just f instead of (f, k) if k is known from context. Then f deg = k. In particular, we often write 0 instead of (0, k) by abuse of notation. Given a shift-graded linear map f : L → M of degree k, given a ∈ Mor(Z) and given ℓ ∈ L^{a}, we often write ℓf instead of ℓf^{a}.

Example 15 We make use of Remark 14. Suppose that Z = Z . A complex is a Z-graded module M, together with a shift-graded linear map d : M → M of degree 1 such that d^{2} = 0. Removing the abusive language of Remark 14 again, we should write (d, 1) in place of d.

So d = (M^{i} ⟶^{d^{i}} M^{i+1})_{i∈Z} .

Moreover, we should write (d, 1)^{2} = (d · d^{[1]}, 2) in place of d^{2}. So in fact, we require

(d · d^{[1]}, 2) = (0, 2), i.e. 0 = d · d^{[1]} = (M^{i} ⟶^{d^{i}·d^{i+1}} M^{i+2})_{i∈Z} , i.e. d^{i} · d^{i+1} = 0 for i ∈ Z.

1.1.3 Tensor products

Let Z = (Z, S, deg) be a grading category. We will not make use of Remark 14 in this §1.1.3.

Definition 16 Suppose given n ∈ Z_{⩾1} .

(1) Suppose given a Z-graded module M_{i} for i ∈ [1, n].

Let ⨂_{i∈[1,n]} M_{i} be the Z-graded module defined by

( ⨂_{i∈[1,n]} M_{i})^{z} := ⨁_{\underline{y} ∈ fact_{n}(z)} ⨂_{i ∈ [1,n]} M_{i}^{y_{i}}

for z ∈ Mor(Z).

We often write M_{1} ⊗ … ⊗ M_{n} := ⨂_{i∈[1,n]} M_{i}.

(2) Suppose given (u_{i})_{i∈[1,n]} , (v_{i})_{i∈[1,n]} ∈ Z^{×n}. Let ⌊(u_{i})_{i} , (v_{i})_{i}⌋ := (−1)^{∑_{1⩽i<j⩽n} u_{i}v_{j}} .

(3) Suppose given Z-graded modules L_{i} and M_{i} for i ∈ [1, n]. Suppose given shift-graded linear maps L_{i} ⟶^{(f_{i} , k_{i})} M_{i} for i ∈ [1, n]. Write k := ∑_{i∈[1,n]} k_{i}. Define the shift-graded linear map

⨂_{i∈[1,n]} L_{i} \underrightarrow{⨂_{i∈[1,n]}(f_{i} , k_{i}) := ( ⨂_{i∈[1,n]}f_{i} , k)} ⨂_{i∈[1,n]} M_{i}

at z ∈ Mor(Z) by

( ⨂_{i∈[1,n]}L_{i})^{z} = ⨁_{\underline{y} ∈ fact_{n}(z)} ⨂_{i ∈ [1,n]} L_{i}^{y_{i}} \underrightarrow{( ⨂_{i∈[1,n]}f_{i})^{z}} ⨁_{\underline{ỹ} ∈ fact_{n}(z[k])} ⨂_{i ∈ [1,n]} M ỹ_{i} ; i = ( ⨂_{i∈[1,n]} M_{i})^{z[k]} ,

mapping an elementary tensor

(ℓ_{i})_{i∈[1,n]}^{⊗} ∈ ⨂_{i ∈ [1,n]} L_{i}^{y_{i}}

to

((ℓ_{i})_{i∈[1,n]}^{⊗})( ⨂_{i∈[1,n]} f_{i})^{z} := ⌊(k_{i})_{i} , (y_{i} deg)_{i}⌋(ℓ_{i}f_{i}^{y_{i}})_{i∈[1,n]}^{⊗} ∈ ⨂_{i ∈ [1,n]} M_{i}^{y_{i}[k_{i}]} .

The sign ⌊(k_{i})_{i} , (ℓ_{i} deg)_{i}⌋ ∈ {−1, +1} is called the Koszul sign ; cf. (2). Note that y_{i} deg = ℓ_{i} deg. Note that in fact, y_{1}[k_{1}] · y_{2}[k_{2}] ⋯ y_{n}[k_{n}] = (y_{1} · y_{2} ⋯ y_{n})[k] = z[k]. We often write (f_{1} , k_{1}) ⊗ … ⊗ (f_{n} , k_{n}) := ⨂_{i∈[1,n]}(f_{i} , k_{i}) .

(3) In Problem 7.(3), we construct a Z-graded module Ṙ such that

Ṙ ⊗ M = M = M ⊗ Ṙ

for a Z-graded module M and, more precisely, (f, k) ⊗ (idṘ , 0) = (f, k) = (idṘ , 0) ⊗ (f, k)

for a shift-graded linear map M ⟶^{(f,k)} N between Z-graded modules M and N. We stipulate that ⨂_{i∈[1,0]} M_{i} := Ṙ and that ⨂_{i∈[1,0]}(f_{i} , k_{i}) = idṘ, in the context of (1, 2). In particular,

M^{⊗k} := ⨂_{i∈[1,k]} M

and

(f, k)^{⊗k} := ⨂_{i∈[1,k]}(f, k)

are defined for k ∈ Z_{⩾0} , where M^{⊗0} = Ṙ and (f, k)^{⊗0} = idṘ.

Example 17 Suppose given n ∈ Z_{⩾1} and (L_{i} ⟶^{(f_{i} , k_{i})} M_{i})_{i∈[1,n]} ∈ Mor((Z-grad)^{×n}).

(0) We have ⌊(1, 4, 5) , (−7, 2, 3)⌋ = (−1)^{1·2+1·3+4·3} = −1.

(1) Suppose n = 1. We have ( ⨂_{i∈[1,1]}L_{i} ⟶^{⨂_{i∈[1,1]}(f_{i} , k_{i})} ⨂_{i∈[1,1]}M_{i}) = (L_{1} ⟶^{(f_{1} , k_{1})} M_{1}).

Note that fact_{1}(z) = {z} for z ∈ Mor(Z), cf. Example 9.(2), and that the Koszul sign is +1 if n = 1.

(2) Suppose n = 2. The shift-graded linear map

( ⨂_{i∈[1,2]} L_{i} ⟶^{⨂_{i∈[1,2]}(f_{i} , k_{i})} ⨂_{i∈[1,2]} M_{i}) = (L_{1} ⊗ L_{2} ⟶^{(f_{1}⊗f_{2} , k)} M_{1} ⊗ M_{2})

of degree k := k_{1} + k_{2} has at z ∈ Mor(Z) the entry

(L_{1} ⊗ L_{2})^{z} = ⨁_{\underline{y} ∈ fact_{2}(z)} L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}} ⟶^{(f_{1}⊗f_{2})^{z}} ⨁_{\underline{ỹ} ∈ fact_{2}(z[k])} M ỹ_{1} ; 1 ⊗ M ỹ_{2} ; 2 = (M_{1} ⊗ M_{2})^{z[k]} ,

mapping an elementary tensor

ℓ_{1} ⊗ ℓ_{2} ∈ L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}}

to

(ℓ_{1} ⊗ ℓ_{2})(f_{1} ⊗ f_{2})^{z} = (−1)^{k_{1}·(ℓ_{2} deg)}(ℓ_{1}f_{1}^{y_{1}} ⊗ ℓ_{2}f_{2}^{y_{2}}) ∈ M_{1}^{y_{1}[k_{1}]} ⊗ M_{2}^{y_{2}[k_{2}]} .

Here, the Koszul sign

⌊(k_{1}, k_{2}) , (ℓ_{1} deg, ℓ_{2} deg)⌋ = (−1)^{k_{1}·(ℓ_{2} deg)}

can be interpreted as being caused by pulling f_{1} , of degree k_{1} , across ℓ_{2} , of degree ℓ_{2} deg.

Consider the case Z = Z. Then z ∈ Z. The map

(L_{1} ⊗ L_{2})^{z} = ⨁_{\begin{matrix}y_{1} , y_{2} ∈ Z, \\ y_{1}+y_{2} = z \end{matrix}}L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}} ⟶^{(f_{1}⊗f_{2})^{z}} ⨁_{\begin{matrix}ỹ_{1} , ỹ_{2} ∈ Z, \\ ỹ_{1}+ỹ_{2} = z+k \end{matrix}}M ỹ_{1} ; 1 ⊗ M ỹ_{2} ; 2 = (M_{1} ⊗ M_{2})^{z+k}

maps

ℓ_{1} ⊗ ℓ_{2} ∈ L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}}

to

(ℓ_{1} ⊗ ℓ_{2})(f_{1} ⊗ f_{2})^{z} = (−1)^{k_{1}·y_{2}}(ℓ_{1}f_{1} ⊗ ℓ_{2}f_{2}) ∈ M_{1}^{y_{1}+k_{1}} ⊗ M_{2}^{y_{2}+k_{2}} .

(3) Suppose n = 3. The shift-graded linear map

( ⨂_{i∈[1,3]} L_{i} ⟶^{⨂_{i∈[1,3]}(f_{i} , k_{i})} ⨂_{i∈[1,3]} M_{i}) = (L_{1} ⊗ L_{2} ⊗ L_{3} ⟶^{(f_{1}⊗f_{2}⊗f_{3} , k)} M_{1} ⊗ M_{2} ⊗ M_{3})

of degree k := k_{1} + k_{2} + k_{3} has at z ∈ Mor(Z) the entry

(L_{1} ⊗ L_{2} ⊗ L_{3})^{z} = ⨁_{\underline{y} ∈ fact_{3}(z)} L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}} ⊗ L_{3}^{y_{3}} ⟶^{(f_{1}⊗f_{2}⊗f_{3})^{z}} ⨁_{\underline{ỹ} ∈ fact_{3}(z[k])} M ỹ_{1} ; 1 ⊗ M ỹ_{2} ; 2 ⊗ M ỹ_{3} ; 3 = (M_{1} ⊗ M_{2} ⊗ M_{3})^{z[k]} ,

mapping an elementary tensor

ℓ_{1} ⊗ ℓ_{2} ⊗ ℓ_{3} ∈ L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}} ⊗ L_{3}^{y_{3}}

to (ℓ_{1} ⊗ ℓ_{2} ⊗ ℓ_{3})(f_{1} ⊗ f_{2} ⊗ f_{3})^{z} = (−1)^{k_{1}·(ℓ_{2} deg +ℓ_{3} deg)+k_{2}·(ℓ_{3} deg)}(ℓ_{1}f_{1}^{y_{1}} ⊗ ℓ_{2}f_{2}^{y_{2}} ⊗ ℓ_{3}f_{3}^{y_{3}})

∈ M_{1}^{y_{1}[k_{1}]} ⊗ M_{2}^{y_{2}[k_{2}]} ⊗ M_{3}^{y_{3}[k_{3}]}

Here, the Koszul sign

⌊(k_{1}, k_{2}, k_{3}) , (ℓ_{1} deg, ℓ_{2} deg, ℓ_{3} deg)⌋ = (−1)^{k_{1}·(ℓ_{2} deg +ℓ_{3} deg)+k_{2}·(ℓ_{3} deg)}

can be interpreted as being caused by pulling f_{1} , of degree k_{1} , across ℓ_{2} ⊗ ℓ_{3} , of degree ℓ_{2} deg +ℓ_{3} deg, and f_{2} , of degree k_{2} , across ℓ_{3} , of degree ℓ_{3} deg. Consider the case Z = Z. Then z ∈ Z. The map

(L_{1} ⊗ L_{2} ⊗ L_{3})^{z} = ⨁_{\begin{matrix}y_{1} , y_{2} , y_{3} ∈ Z, \\ y_{1}+y_{2}+y_{3} = z \end{matrix}}L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}} ⊗ L_{3}^{y_{3}} ⟶^{(f_{1}⊗f_{2}⊗f_{3})^{z}} ⨁_{\begin{matrix}ỹ_{1} , ỹ_{2} , ỹ_{3} ∈ Z, \\ ỹ_{1}+ỹ_{2}+ỹ_{3} = z+k \end{matrix}}M ỹ_{1} ; 1 ⊗ M ỹ_{2} ; 2 ⊗ M ỹ_{3} ; 3 = (M_{1} ⊗ M_{2} ⊗ M_{3})^{z+k}

maps

ℓ_{1} ⊗ ℓ_{2} ⊗ ℓ_{3} ∈ L_{1}^{y_{1}} ⊗ L_{2}^{y_{2}} ⊗ L_{3}^{y_{3}}

to

(ℓ_{1}⊗ℓ_{2}⊗ℓ_{3})(f_{1}⊗f_{2}⊗f_{3})^{z} = (−1)^{k_{1}·(y_{2}+y_{3})+k_{2}·y_{3}}(ℓ_{1}f_{1}⊗ℓ_{2}f_{2}⊗ℓ_{3}f_{3}) ∈ M_{1}^{y_{1}+k_{1}} ⊗M_{2}^{y_{2}+k_{2}} ⊗M_{3}^{y_{3}+k_{3}} .

Example 18 We consider the tensor product of two complexes; cf. Example 15. Let M be a complex with differential (d, 1). Let M̃ be a complex with differential (d̃, 1). Then the Z-graded module M ⊗ M̃ has at position z ∈ Z the entry

⨁_{i, j ∈ Z, i+j=z} M^{i} ⊗ M̃^{j} .

The entry at i + j of the graded linear map (D, 1) := (d, 1) ⊗ (id, 0) + (id, 0) ⊗ (d̃, 1) of degree 1 maps the elementary tensor

m ⊗ m̃ ∈ M^{i} ⊗ M̃^{j}

to

\begin{aligned}(m ⊗ m̃)D & = & (−1)^{1·j}md^{i} ⊗ m̃ id +(−1)^{0·j}m id ⊗m̃d̃^{j} \\ & = & (−1)^{j}md^{i} ⊗ m̃ + m ⊗ m̃d̃^{j} . \end{aligned}

So entry at i + j of the graded linear map (D, 1)^{2} = (D · D^{[}1], 2) of degree 2 maps it to

((m ⊗ m̃)D)D^{[1]}

= ((−1)^{j}md^{i} ⊗ m̃ + m ⊗ m̃d̃^{j})D^{[}1]
= (−1)^{j+j}md^{i}d^{i+1} ⊗ m̃ + (−1)^{j}md^{i} ⊗ m̃d̃^{j} + (−1)^{j+1}md^{i} ⊗ m̃d̃^{j} + m ⊗ m̃d̃^{j}d̃^{j+1}
= 0 + (−1)^{j}md^{i} ⊗ m̃d̃^{j}(1 − 1) + 0
= 0 .

So (D, 1)^{2} = 0. Hence M ⊗ M̃, with differential (D, 1), is a complex.

This would not have been the case without inserting a sign such as the Koszul sign.

1.2 A_{∞}-algebras and A_{∞}-categories

Let Z = (Z, S, deg) be a grading category; cf. Definition 5.

Henceforth, we make use of Remark 14.

Definition 19 Suppose given n ∈ [0, ∞].

(1) Suppose given a Z-graded module A.

Suppose given a shift-graded linear map m_{k}^{A} = (m_{k}^{A} , 2 − k) : A^{⊗k} → A of degree 2 − k for k ∈ [1, n] ∩ Z.

Then A = (A, (m_{k}^{A})_{k∈[1,n]∩Z}) is a pre-A_{n}-algebra (over Z).

A pre-A_{n}-algebra A = (A, (m_{k}^{A})_{k∈[1,n]∩Z}) is an A_{n}-algebra (over Z) if the Stasheff equation

0 = ∑_{\begin{matrix}(r,s,t)∈Z_{⩾0}×Z_{⩾1}×Z_{⩾0} \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m_{s}^{A} ⊗ id^{⊗t}) · m_{r+1+t}^{A}

holds for k ∈ [1, n] ∩ Z.

Note that each summand of the right-hand side is a shift-graded linear map from A^{⊗k} to A of degree 3 − k.

We often abbreviate A = (A, (m_{k})_{k}) = (A, (m_{k}^{A})_{k}) := (A, (m_{k}^{A})_{k∈[1,n]∩Z}).

Sometimes, the tuple (m_{k})_{k} is referred to as an A_{n}-structure on the Z-graded module A. An entry m_{k} of this tuple is sometimes referred to as kth shift-graded linear multiplication map.

We often abbreviate the condition (r, s, t) ∈ Z_{⩾0} ×Z_{⩾1} ×Z_{⩾0} on the indexing triples to (r, s, t) ⩾ (0, 1, 0).

(2) Suppose given Z-graded modules Ã and A.

Suppose given a shift-graded linear map f_{k} = (f_{k} , 1 − k) : Ã^{⊗k} → A of degree 1 − k for k ∈ [1, n] ∩ Z.

Then f = (f_{k})_{k∈[1,n]∩Z} : Ã → A is a pre-A_{n}-morphism (over Z).

Suppose given A_{n}-algebras Ã = (Ã, (mÃ ; k)_{k∈[1,n]∩Z}) and A = (A, (m_{k}^{A})_{k∈[1,n]∩Z}).

A pre-A_{n}-morphism f = (f_{k})_{k∈[1,n]∩Z} : Ã → A is an A_{n}-morphism or a morphism of A_{n}-algebras (over Z) if the Stasheff equation for morphisms

∑_{\begin{matrix}(r,s,t)∈Z_{⩾0}×Z_{⩾1}×Z_{⩾0} \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗mÃ ; s ⊗ id^{⊗t}) · f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]} ∈ Z_{⩾1}^{×r} \\ ∑_{j∈[1,r]} i_{j} = k \end{matrix}}⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋( ⨂_{j∈[1,r]} f_{i_{j}}) · m_{r}^{A}

holds for k ∈ [1, n] ∩ Z.

Note that each summand of the left- and of the right-hand side is a shift-graded linear map from Ã^{⊗k} to A of degree 2 − k.

We often abbreviate (f_{k})_{k} = (f_{k})_{k∈[1,n]∩Z} .

We often abbreviate the condition (r, s, t) ∈ Z_{⩾0} ×Z_{⩾1} ×Z_{⩾0} on the indexing triples to (r, s, t) ⩾ (0, 1, 0) and the condition (i_{j})_{j∈[1,r]} ∈ Z_{⩾1}^{×r} on the indexing tuples to (i_{j})_{j} ⩾ (1)_{j}.

Remark 20 Suppose given 1 ⩽ ℓ ⩽ n ⩽ ∞.

(1) Given an A_{n}-algebra (A, (m_{k}^{A})_{k∈[1,n]∩Z}), we get an A_{ℓ}-algebra (A, (m_{k}^{A})_{k∈[1,ℓ]∩Z}).

(2) Given an A_{n}-morphism (f_{k})_{k∈[1,n]∩Z} from an A_{n}-algebra (Ã, (mÃ ; k)_{k∈[1,n]∩Z}) to an A_{n}-algebra (A, (m_{k}^{A})_{k∈[1,n]∩Z}), we get an A_{ℓ}-morphism (f_{k})_{k∈[1,ℓ]∩Z} from

(Ã, (mÃ ; k)_{k∈[1,ℓ]∩Z}) to (A, (m_{k}^{A})_{k∈[1,ℓ]∩Z}) .

Example 21

We consider the Stasheff equation from Definition 19.(1) for an A_{n}-algebra A = (A, (m_{k}^{A})_{k}) = (A, (m_{k})_{k}) for k ∈ [1, 3], supposing n ⩾ k.

(1) For k = 1, the Stasheff equation reads

0 = m_{1} · m_{1} .

So in case Z = Z, the graded module A is a complex with differential m_{1} .

(2) For k = 2, the Stasheff equation reads

0 = −(id ⊗m_{1}) · m_{2} − (m_{1} ⊗ id) · m_{2} + m_{2} · m_{1} .

In case Z = Z, we obtain for a, b ∈ A^{0}

((a ⊗ b)m_{2})m_{1} = (a ⊗ bm_{1})m_{2} + (am_{1} ⊗ b)m_{2} .

Interpreting m_{1} as differential and m_{2} as multiplication, this is a product rule for the differential, often called the Leibniz rule.

(3) For k = 3, the Stasheff equation reads

0 = (m_{1} ⊗ id^{⊗2}) · m_{3} + (id ⊗m_{1} ⊗ id) · m_{3} + (id^{⊗2} ⊗m_{1}) · m_{3} + (m_{2} ⊗ id) · m_{2} − (id ⊗m_{2}) · m_{2} + m_{3} · m_{1} .

In case Z = Z, we obtain for a, b, c ∈ A^{0}

\begin{matrix}(a ⊗ (b ⊗ c)m_{2})m_{2} − ((a ⊗ b)m_{2} ⊗ c) · m_{2} \\ = (am_{1} ⊗ b ⊗ c)m_{3} + (a ⊗ bm_{1} ⊗ c)m_{3} + (a ⊗ b ⊗ cm_{1})m_{3} + ((a ⊗ b ⊗ c)m_{3})m_{1} \end{matrix}

Interpreting m_{2} as multiplication, we observe that this multiplication is associative if m_{3} = 0 or if m_{1} = 0.

We do not claim that associativity of m_{2} entails m_{3} = 0.

Example 22

We consider the Stasheff equation for morphisms from Definition 19.(2) for a morphism f = (f_{k})_{k} : Ã → A of A_{n}-algebras for k ∈ [1, 3], supposing n ⩾ k.

We consider the conditions of Definition 19.(2) for k ∈ [1, 3], supposing n ⩾ k.

(1) For k = 1, we obtain the condition

mÃ ; 1 · f_{1} = f_{1} · m_{1}^{A} .

In case Z = Z, we obtain that f_{1} is a morphism of complexes from Ã, having differential mÃ ; 1 , to A, having differential m_{1}^{A}.

(2) For k = 2, we obtain the condition

\begin{matrix}−(id ⊗mÃ ; 1) · f_{2} − (mÃ ; 1 ⊗ id) · f_{2} + mÃ ; 2 · f_{1} \\ = f_{2} · m_{1}^{A} + (f_{1} ⊗ f_{1}) · m_{2}^{A} . \end{matrix}

(3) For k = 3, we obtain the condition

(mÃ ; 1 ⊗ id^{⊗2}) · f_{3} + (id ⊗mÃ ; 1 ⊗ id) · f_{3} + (id^{⊗2} ⊗mÃ ; 1) · f_{3} + (mÃ ; 2 ⊗ id) · f_{2} − (id ⊗mÃ ; 2) · f_{2} + mÃ ; 3 · f_{1}

= f_{3} · m_{1}^{A} + (f_{1} ⊗ f_{2}) · m_{2}^{A} − (f_{2} ⊗ f_{1}) · m_{2}^{A} + (f_{1} ⊗ f_{1} ⊗ f_{1}) · m_{3}^{A} .

Definition 23 Let n ∈ [2, ∞]. Suppose given an A_{n}-algebra A = (A, (m_{k}^{A})_{k∈[1,n]∩Z}). Then A is called unital, if for X ∈ Ob Z, there exists a neutral element 1_{A,X} = 1_{X} ∈ A^{id^{X}} such that (1, 2) hold.

(1) We have (a ⊗ 1_{X})m_{2}^{A} = a for z ∈ Mor(Z) with zt_{Z} = X and a ∈ A^{z}.

(2) We have (1_{X} ⊗ b)m_{2}^{A} = b for w ∈ Mor(Z) with ws_{Z} = X and b ∈ A^{w}.

Then 1_{X} is uniquely determined, for given an element c ∈ A^{id^{X}} having properties (1, 2), then 1_{X} = (1_{X} ⊗ c)m_{2}^{A} = c.

Note that (1_{X}⊗1_{X})m_{2}^{A}m_{1}^{A} =^{1.} 1_{X}m_{1}^{A} =^{2.} (1_{X}m_{1}^{A}⊗1_{X})m_{2}^{A}+(1_{X}⊗1_{X}m_{1}^{A})m_{2}^{A} = 1_{X}m_{1}^{A}+1_{X}m_{1}^{A}

for X ∈ Ob Z, whence 1_{X}m_{1}^{A} = 0 ; cf. Example 21.(2).

Definition 24 Let n ∈ [2, ∞]. Suppose given unital A_{n}-algebras Ã and A. Suppose given an A_{n}-morphism Ã ⟶^{f} A, so f = (f_{k})_{k∈[1,n]∩Z} . Then f is called unital, if for X ∈ Ob Z, we have 1_{Ã,X}f_{1} = 1_{A,X} .

Example 25 Suppose given n ∈ [3, ∞] . Suppose given a unital A_{n}-algebra A = (A, (m_{k})_{k}). Suppose that A^{z} = 0 for z ∈ Mor(Z) with deg z ∈ Z ∖ {0}. Then m_{k} = 0 for k ∈ [1, n] ∖ {2}. In fact, the shift-graded linear map m_{k} = (m_{k} , 2 − k) actually maps from A^{⊗k} to A^{[2−k]}. So given z ∈ Mor(Z) and \underline{y} ∈ fact_{k}(z) and

a_{1} ⊗ … ⊗ a_{k} ∈ A^{y_{1}} ⊗ … ⊗ A^{y_{k}} ,

its image is

(a_{1} ⊗ … ⊗ a_{k})m_{k} ∈ A^{z[2−k]} .

In order that this image be nonzero, we need that on the one hand, a_{i} is nonzero for i ∈ [1, k], so necessarily y_{i} deg = 0 for i ∈ [1, k]. On the other hand, we must necessarily have z[2 − k] deg = 0. But

0 = z[2 − k] deg = z deg +2 − k = ( ∑_{i∈[1,k]} y_{i}) + 2 − k = 2 − k ,

so k = 2.

In particular, interpreting m_{2} as multiplication, it is associative; cf. Example 21.(3).

Case Z = Z. Then A^{0}, together with the multiplication

\begin{matrix}A^{0} & ⊗ & A^{0} & → & A^{0} \\ a & ⊗ & b & ↦ & a · b := (a ⊗ b)m_{2} \end{matrix}

is an algebra. The element 1_{Z} ∈ A^{id^{Z}} = A^{0} is neutral with respect to multiplication ; cf. Definition 23.

Case Z = Z × I^{×2} for a set I; cf. Example 1.(4). Then we have a linear category A^{0} with Ob(A^{0}) = I and _{A^{0}}(i, j) = A^{(0,(i,j))} for i, j ∈ I. Its composition is given by

\begin{matrix}A^{(0,(i,j))} & ⊗ & A^{(0,(j,k))} & → & A^{(0,(i,k))} \\ a & ⊗ & b & ↦ & a · b := (a ⊗ b)m_{2} \end{matrix}

for i, j, k ∈ I. Given (Z, i) ∈ {Z} × I = Ob(Z × I^{×2}), the element 1_{(Z,i)} ∈ A^{id^{(Z,i)}} = A^{(0,(i,i))} is neutral with respect to composition ; cf. Definition 23.

Example 26 Suppose given n ∈ [3, ∞].

Suppose given unital A_{n}-algebras Ã and A.

Suppose given a unital A_{n}-morphism Ã ⟶^{f} A, so f = (f_{k})_{k∈[1,n]∩Z} .

Suppose that Ã^{z} = 0 and A^{z} = 0 for z ∈ Mor(Z) with deg z ∈ Z ∖ {0}. So mÃ ; k = 0 and m_{k}^{A} = 0 for k ∈ [1, n] ∖ {2}; cf. Example 25.

Moreover, f_{k} = 0 for k ∈ [1, n] ∖ {1}. In fact, the shift-graded linear map f_{k} = (f_{k}, 1 − k) actually maps from Ã^{⊗k} to A^{[1−k]}. So given z ∈ Mor(Z) and \underline{y} ∈ fact_{k}(z) and

a_{1} ⊗ … ⊗ a_{k} ∈ A^{y_{1}} ⊗ … ⊗ A^{y_{k}} ,

its image is

(a_{1} ⊗ … ⊗ a_{k})f_{k} ∈ A^{z[1−k]} .

In order that this image be nonzero, we need that on the one hand, a_{i} is nonzero for i ∈ [1, k], so necessarily y_{i} deg = 0 for i ∈ [1, k]. On the other hand, we must necessarily have z[1 − k] deg = 0. But

0 = z[1 − k] deg = z deg +1 − k = ( ∑_{i∈[1,k]} y_{i}) + 1 − k = 1 − k ,

so k = 1.

Now Ã ⟶^{f_{1}} A satisfies mÃ ; 2 · f_{1} = (f_{1} ⊗ f_{1}) · m_{2}^{A}; cf. Example 22.(2).

Case Z = Z. Then Ã^{0} and A^{0} are algebras; cf. Example 25. Using the multiplication notation from there, we obtain

(ã · b̃)f_{1} = (ã ⊗ b̃)(mÃ ; 2 · f_{1}) = (ã ⊗ b̃)(f_{1} ⊗ f_{1}) · m_{2}^{A} = ãf_{1} · b̃f_{1}

for ã, b̃ ∈ Ã^{0}. Moreover, we have

1_{Ã,Z}f_{1} = 1_{A,Z} .

So f_{1} is a morphism of algebras from Ã^{0} to A^{0}.

Case Z = Z × I^{×2} for a set I; cf. Example 1.(4). Then Ã^{0} and A^{0} are linear categories; cf. Example 25. Using the composition notation from there, we obtain

(ã · b̃)f_{1} = (ã ⊗ b̃)(mÃ ; 2 · f_{1}) = (ã ⊗ b̃)(f_{1} ⊗ f_{1}) · m_{2}^{A} = ãf_{1} · b̃f_{1}

for i, j, k ∈ I, for a ∈ Ã^{(0,(i,j))} and a ∈ Ã^{(0,(j,k))}. Moreover, we have

1Ãf_{1} = 1_{A} .

So f_{1} is a linear functor from Ã^{0} to A^{0}.

Exceptionally, f_{1} is written on the right, i.e. naturally.

Definition 27 Recall that Z is a grading category.

(1) An A_{∞}-algebra over Z is called a classical A_{∞}-algebra.

(2) Suppose given a set I.

A unital A_{∞}-algebra A over Z × I^{×2} is called an A_{∞}-category with set of objects Ob(A) = I; cf. Example 1.(4).

(3) A unital A_{∞}-algebra A = (A, (m_{k})_{k∈Z_{⩾1}}) over Z with m_{k} = 0 for k ⩾ 3 is called a differential graded algebra (over Z). Cf. Problem 11.

(4) A unital A_{∞}-algebra A = (A, (m_{k})_{k∈Z_{⩾1}}) over Z with m_{k} = 0 for k ⩾ 3 is called a classical differential graded algebra.

(5) Suppose given a set I.

A unital A_{∞}-algebra A = (A, (m_{k})_{k∈Z_{⩾1}}) over Z × I^{×2} with m_{k} = 0 for k ⩾ 3 is called a differential graded category with set of objects I; cf. Example 1.(4).

(6) A unital A_{∞}-algebra A = (A, (m_{k})_{k∈Z_{⩾1}}) over Z with m_{1} = 0 is called minimal.

1.3 The regular differential graded category for complexes

Suppose given an algebra B.

Suppose given N ∈ Z_{⩾1} . Suppose given complexes X_{s} ∈ Ob(C(B -Mod)) for s ∈ [1, N], where X_{s} carries the differential d_{s} = (X_{s}^{i} ⟶^{d_{s}^{i}} X_{s}^{i+1})_{i}. Write \underline{X} := (X_{s})_{s∈[1,N]} . Let Z := Z × [1, N]^{×2}. Write

Hom_{B}^{j}(X_{s} , X_{t}) := { (f^{i})_{i∈Z} : X_{s}^{i} ⟶^{f^{i}} X_{t}^{i+j} is a B-linear map for i ∈ Z }

for s, t ∈ [1, N] and j ∈ Z.

In general, the inclusion _{C(B -Mod)}(X_{s} , X_{t}) ⊆ Hom_{B}^{0}(X_{s} , X_{t}) is strict. The definition of

Hom_{B}^{j}(X_{s} , X_{t}) does not involve the differentials of X_{s} and of X_{t}.

We shall construct the regular differential graded algebra Hom_{B}(\underline{X}) of \underline{X} over Z on the set of objects [1, N]. As a Z-graded module, define Hom_{B}(\underline{X}) by letting

Hom_{B}(\underline{X})^{(j,(s,t))} := Hom_{B}^{j}(X_{s} , X_{t})

for (j, (s, t)) ∈ Mor(Z). Let

Hom_{B}(\underline{X}) ⟶^{m_{1}^{Hom_{B} (\underline{X})}} Hom_{B}(\underline{X})

be defined at (j, (s, t)) ∈ Mor(Z) by

Hom_{B}^{j}(X_{s} , X_{t}) ⟶^{m_{1}^{Hom_{B} (\underline{X})}} Hom_{B}^{j+1}(X_{s} , X_{t})
(f^{i})_{i}  ↦  (f^{i}d_{t}^{i+j} − (−1)^{j}d_{s}^{i}f^{i+1})_{i} .
\xymatrix{X_{s}^{i} \ar[r]^{f^{i}} \ar[d]_{d_{s}^{i}} & X_{t}^{i+j} \ar[d]^{d_{t}^{i+j}} \\ X_{s}^{i+1} \ar[r]^{f^{i+1}} & X_{t}^{i+j+1}}

Let

Hom_{B}(\underline{X})^{⊗2} ⟶^{m_{2}^{Hom_{B} (\underline{X})}} Hom_{B}(\underline{X})

be defined at (j, (s, t)) ∈ Mor(Z), on the summand belonging to

((k, (s, u)), (ℓ, (u, t))) ∈ fact_{2} ((j, (s, t))) ,

i.e. k + ℓ = j and u ∈ [1, N], by

Hom_{B}^{k}(X_{s} , X_{u}) ⊗ Hom_{B}^{ℓ}(X_{u} , X_{t}) ⟶^{m_{2}^{Hom_{B} (\underline{X})}} Hom_{B}^{j}(X_{s} , X_{t})
(f^{i})_{i} ⊗ (g^{i})_{i}  ↦  (f^{i}g^{i+k})_{i} .
\xymatrix{X_{s}^{i} \ar[r]^{f^{i}} & X_{u}^{i+k} \ar[r]^{g^{i+k}} & X_{t}^{i+k+ℓ}}

Lemma 28 Recall that Z = Z × [1, N]^{×2}. Recall that X_{s} ∈ C(B -Mod). Consider the Z-graded module Hom_{B}(\underline{X}) and the shift-graded morphisms

Hom_{B}(\underline{X}) ⟶^{m_{1}^{Hom_{B} (\underline{X})}} Hom_{B}(\underline{X})

of degree 1 and

Hom_{B}(\underline{X})^{⊗2} ⟶^{m_{2}^{Hom_{B} (\underline{X})}} Hom_{B}(\underline{X})

of degree 0 constructed above. Let m_{k}^{Hom_{B}(\underline{X})} := 0, as a shift-graded linear map of degree 2 − k from Hom_{B}(\underline{X})^{⊗k} to Hom_{B}(\underline{X}), for k ∈ Z_{⩾3} . Then (Hom_{B}(\underline{X}), (m_{k}^{Hom_{B}(\underline{X})})_{k∈Z_{⩾1}}) is a differential graded category on the set of objects [1, N] ; cf. Definition 27.(5).

Proof. We have to show the Stasheff equation for k ∈ [1, 3] and the existence of neutral elements; cf. Problem 11. Write m_{k} := m_{k}^{Hom_{B}(\underline{X})} for k ∈ Z_{⩾1} . Case k = 1. We have to show that m_{1} · m_{1} =^{!} 0 ; cf. Example 21.(1). Given (j, (s, t)) ∈ Mor(Z) and (f_{i})_{i} ∈ Hom_{B}^{j}(X_{s} , X_{t}), we obtain

((f_{i})_{i})(m_{1} · m_{1})

= ((f^{i}d_{t}^{i+j} − (−1)^{j}d_{s}^{i}f^{i+1})_{i})m_{1}
= ((f^{i}d_{t}^{i+j} − (−1)^{j}d_{s}^{i}f^{i+1})d_{t}^{i+j+1} − (−1)^{j+1}d_{s}^{i}(f^{i+1}d_{t}^{i+j+1} − (−1)^{j}d_{s}^{i+1}f^{i+2}))_{i}
= (f^{i}d_{t}^{i+j}d_{t}^{i+j+1} − (−1)^{j}d_{s}^{i}f^{i+1}d_{t}^{i+j+1} − (−1)^{j+1}d_{s}^{i}f^{i+1}d_{t}^{i+j+1} + (−1)^{j+1}(−1)^{j}d_{s}^{i}d_{s}^{i+1}f^{i+2}))_{i}
= (0)_{i} .

Case k = 2. We have to show that m_{2} ⊗ m_{1} =^{!} (id ⊗m_{1}) · m_{2} + (m_{1} ⊗ id) · m_{2} ; cf. Example 21.(2). Given (j, (s, t)) ∈ Mor(Z) and ((k, (s, u)), (ℓ, (u, t))) ∈ fact_{2} ((j, (s, t))), i.e. k + ℓ = j and u ∈ [1, N], and (f^{i})_{i} ∈ Hom_{B}^{k}(X_{s} , X_{u}) and (g^{i})_{i} ∈ Hom_{B}^{ℓ}(X_{u} , X_{t}), we obtain

((f^{i})_{i} ⊗ (g^{i})_{i})(m_{2} ⊗ m_{1})

= ((f^{i}g^{i+k})_{i})m_{1}
= (f^{i}g^{i+k}d_{t}^{i+j} − (−1)^{j}d_{s}^{i}f^{i+1}g^{i+1+k})_{i}

and

((f^{i})_{i} ⊗ (g^{i})_{i})((id ⊗m_{1}) · m_{2} + (m_{1} ⊗ id) · m_{2})
=^{Koszul} ((f^{i})_{i} ⊗ ((g^{i})_{i})m_{1})m_{2} + (−1)^{ℓ}(((f^{i})_{i})m_{1} ⊗ (g^{i})_{i})m_{2}
= ((f^{i})_{i} ⊗ (g^{i}d_{t}^{i+ℓ} − (−1)^{ℓ}d_{u}^{i}g^{i+1})_{i})m_{2} + (−1)^{ℓ}((f^{i}d_{u}^{i+k} − (−1)^{k}d_{s}^{i}f^{i+1})_{i} ⊗ (g^{i})_{i})m_{2}
= (f^{i}g^{i+k}d_{t}^{i+k+ℓ} − (−1)^{ℓ}f^{i}d_{u}^{i+k}g^{i+k+1})_{i} + (−1)^{ℓ}(f^{i}d_{u}^{i+k}g^{i+k+1} − (−1)^{k+ℓ}d_{s}^{i}f^{i+1}g^{i+k+1})_{i}
= (f^{i}g^{i+k}d_{t}^{i+k+ℓ} − (−1)^{k+ℓ}d_{s}^{i}f^{i+1}g^{i+k+1})_{i} ,

which is the same. Case k = 3. We have to show that (m_{2} ⊗ id) · m_{2} =^{!} (id ⊗m_{2}) · m_{2} ; cf. Example 21.(3). Given (j, (s, t)) ∈ Mor(Z) and ((k, (s, u)), (ℓ, (u, v)), (p, (v, t))) ∈ fact_{3} ((j, (s, t))), i.e. k + ℓ + p = j and u, v ∈ [1, N], and (f^{i})_{i} ∈ Hom_{B}^{k}(X_{s} , X_{u}) and (g^{i})_{i} ∈ Hom_{B}^{ℓ}(X_{u} , X_{v}) and (h^{i})_{i} ∈ Hom_{B}^{ℓ}(X_{v} , X_{t}), we obtain

((f^{i})_{i} ⊗ (g^{i})_{i} ⊗ (h^{i})_{i})((m_{2} ⊗ id) · m_{2})

=^{Koszul} (((f^{i})_{i} ⊗ (g^{i})_{i})m_{2} ⊗ (h^{i})_{i})m_{2}
= ((f^{i}g^{i+k})_{i} ⊗ (h^{i})_{i})m_{2}
= (f^{i}g^{i+k}h^{i+k+ℓ})_{i}

and

((f^{i})_{i} ⊗ (g^{i})_{i} ⊗ (h^{i})_{i})((id ⊗m_{2}) · m_{2})
=^{Koszul} ((f^{i})_{i} ⊗ ((g^{i})_{i} ⊗ (h^{i})_{i})m_{2})m_{2}
= ((f^{i})_{i} ⊗ (g^{i}h^{i+ℓ})_{i})m_{2}
= (f^{i}g^{i+k}h^{i+k+ℓ})_{i} ,

which is the same. We have to show the existence of neutral elements, i.e. that Hom_{B}(\underline{X}) is unital; cf. Definition 23. Given (Z, s) ∈ Ob(Z), let

1_{s} := 1_{(Z,s)} := (id_{X_{s}^{i}})_{i} .

Given (j, (s, t)) ∈ Mor(Z) and (f_{i})_{i} ∈ Hom_{B}^{j}(X_{s} , X_{t}), we obtain (j, (s, t))t_{Z} = (Z, t) and ((f_{i})_{i} ⊗ 1_{(Z,t)})m_{2} = (f_{i} · id_{X_{t}^{i+j}})_{i} = (f_{i})_{i} ,

and we obtain (j, (s, t))s_{Z} = (Z, s) and (1_{(Z,s)} ⊗ (f_{i})_{i})m_{2} = (id_{X_{s}^{i}} ·f^{i})_{i} = (f^{i})_{i} .

1.4 Cohomology

Let Z be a grading category.

Definition 29 Let n ∈ [1, ∞].

(1) Suppose given an A_{n}-algebra A over Z. Let ZA := Kern(m_{1}) be the Z-graded module of cycles.

Let BA := Im(m_{1}) be the Z-graded module of boundaries.

Note that BA ⊆ ZA since m_{1}^{2} = 0 ; cf. Example 21.(1).

Let HA := (ZA)/(BA) be the Z-graded cohomology module of A.

Specifically, we have, at z ∈ Mor(Z),

(ZA)^{z} = Kern(A^{z} ⟶^{m_{1}^{A}} A^{z[1]})
(BA)^{z} = Im(A^{z[−1]} ⟶^{m_{1}^{A}} A^{z}) .

Note that (BA)^{z} ⊆ (ZA)^{z}; cf. Example 21.(1).

(2) Suppose given a morphism Ã ⟶^{f} A of A_{n}-algebras. We shall define a shift-graded linear map

HÃ ⟶^{Hf} HA

of degree 0. At z ∈ Mor(Z), it is given by

\begin{matrix}(HÃ)^{z} & ⟶^{(Hf)^{z}} & (HÃ)^{z} \\ a + (BÃ)^{z} & ↦ & af_{1} + (BA)^{z} \end{matrix}

This is a welldefined linear map, since f_{1} maps (BÃ)^{z} to (BA)^{z} as well as (ZÃ)^{z} to (ZA)^{z}, because given a′ ∈ Ã^{z[−1]}, we get

a′mÃ ; 1f_{1} = a′f_{1}m_{1}^{A} ∈ (BA)^{z} ;

cf. Example 22.(1).

Sometimes, we also write Hf_{1} := Hf.

(3) A morphism Ã ⟶^{f} A of A_{n}-algebras is called a quasiisomorphism if Hf is an isomorphism.

Since we do not know yet how to compose A_{n}-morphisms, we do not have a category of A_{n}-algebras at our disposal. Hence, at this point, we cannot decide whether H is a functor from the category of A_{n}-algebras over Z to Z-grad. Cf. Problem 23.(7) below.

Our aim is to show the Theorem of Kadeishvili, Theorem 50 below, which, in case R is a field, will establish the existence of a minimal A_{n}-structure on A and at the same time a quasiisomorphism from HA to A. This theorem seems to be hard to obtain by a direct calculation, though. We will make a detour, reinterpret Stasheff equations as a codifferential condition on a tensor coalgebra, in order to obtain an understandable proof. In the following Remark 30, we illustrate the first two steps towards Kadeishvili.

Remark 30 Let n ∈ [1, ∞]. Suppose given an A_{n}-algebra A over Z.

Denote by BA ⟶^{ι̃} ZA ⟶^{ι} A the inclusion morphisms.

Denote by ZA ⟶^{ρ} HA the residue class morphism. Note that BA = Kern(ρ).

In particular, given a morphism Ã ⟶^{f} A of A_{n}-algebras, we get the following commutative diagram.

\xymatrix{Ã \ar[r]^{f} & A \\ ZÃ \ar[r]^{f|_{ZA}^{ZÃ}} \ar[u]^{ι} \ar[d]_{ρ} & ZA \ar[u]_{ι} \ar[d]^{ρ} \\ HÃ \ar[r]^{Hf = Hf_{1}} & HA}

If R is a field, we may choose a shift-graded linear map ZA ⟵^{σ} HA of degree 0 such that σρ = id_{HA}; cf. Problem 15.(2). If R is a field, we may choose a shift-graded linear map A ⟵^{τ} BA of degree −1 such that τ(m_{1}|^{BA}) = id_{HA}; cf. Problem 15.(2). Since (id_{ZA} −ρ · σ)ρ = ρ − ρ · σρ = ρ − ρ = 0, there exists a unique shift-graded linear map ZA ⟶^{ν̃} BA of degree 0 such that ν̃ · ι̃ = id_{ZA} −ρ · σ; cf. Problem 15.(1). Write ν := ν̃τ. So

ν · m_{1} = ν̃ · τ · (m_{1}|^{BA}) · ι̃· ι = ν̃ · ι̃· ι = (id_{ZA} −ρ · σ) · ρ .
\xymatrix{ &  & A \\ A \ar[urr]^{m_{1}} \ar[r]^{m_{1}|^{BA}} & BA \ar[r]^{ι̃} \ar@{.>}[l]^{τ} & ZA \ar[u]_{ι} \ar[r]^{ρ} & HA \ar@{.>}[l]^{σ} \\  &  & ZA \ar[u]_{id_{ZA} −ρ·σ} \ar[ul]^{ν̃} \ar@{.>}[ull]^{ν}}

Here the existence of the shift graded linear maps written with dotted arrows is only ensured if R is a field.

Remark 31 Suppose R to be a field. Suppose given an A_{3}-algebra A over Z. We will construct a minimal A_{3}-structure (m̃_{1}, m̃_{2}, m̃_{3}) on HA and a quasiisomorphism (q_{1}, q_{2}, q_{3}) of A_{3}-algebras from HA to A. Step 1. For this step, we will only need A as an A_{1}-algebra. Let m̃_{1} = 0. Let q_{1} := σ · ι. We have

m̃_{1} · m̃_{1} = 0.

Hence the Stasheff equation at k = 1 holds; cf. Example 21.(1).

We have

m̃_{1} · q_{1} = 0 = q_{1} · m_{1} .

Hence the Stasheff equation for morphisms at k = 1 holds; cf. Example 22.(1). Since we have the commutative diagram

[figure: HA; q_{1}; A; ZHA; σ; ZA; ι; ρ; HHA; HA ,]

we have Hq_{1} = id_{HA}, which is an isomorphism. So q = (q_{1}, q_{2}, q_{3}), once constructed as a morphism of A_{3}-algebras, will be a quasiisomorphism; cf. Definition 29.(2, 3). Step 2. For this step, we will only need A as an A_{2}-algebra. By Example 21.(2), we get

ι^{⊗2} · m_{2} · m_{1} = ι^{⊗2}(m_{1} ⊗ id + id ⊗m_{1})
= (ιm_{1} ⊗ ι + ι ⊗ ιm_{1})
= 0 .

Thus by Problem 15.(1), we get a unique shift-graded linear map m̌_{2} : (ZA)^{⊗2} → ZA of degree 0 such that the following quadrangle commutes.

\xymatrix{A^{⊗2} \ar[r]^{m_{2}} & A \\ (ZA)^{⊗2} \ar[r]^{m̌_{2}} \ar[u]^{ι^{⊗2}} & ZA \ar[u]_{ι}}

We claim that ((m_{1}|^{ZA}) ⊗ id_{ZA}) · m̌_{2} · ρ =^{!} 0 and that (id_{ZA} ⊗(m_{1}|^{ZA})) · m̌_{2} · ρ =^{!} 0. We prove the first equation. The second then follows by an analogous reasoning. Given z ∈ Mor(Z) and (u, v) ∈ fact_{2}(z[−1]) and a ∈ A^{u} and ã ∈ (ZA)^{v}, we have a ⊗ ã ∈ (A^{⊗2})^{z[−1]} and obtain

(a ⊗ ã)((m_{1}|^{ZA}) ⊗ id_{ZA})m̌_{2} · ρ = (a ⊗ ã)(m_{1} ⊗ id)m_{2} + (BA)^{z}
=^{Ex. 21.(2)} −(a ⊗ ã)(id ⊗m_{1})m_{2} + (a ⊗ ã)m_{2} · m_{1} + (BA)^{z}
= −(a ⊗ ãm_{1}) + (a ⊗ ã)m_{2} · m_{1} + (BA)^{z}
= ((a ⊗ ã)m_{2})m_{1} + (BA)^{z}
= 0 .

This proves the claim. So by Problem 16.(2), we obtain a unique shift-graded linear map m̂_{2} : (HA)^{⊗2} → HA of degree 0 such that the following quadrangle commutes.

\xymatrix{(ZA)^{⊗2} \ar[r]^{m̌_{2}} \ar[d]_{ρ^{⊗2}} & ZA \ar[d]^{ρ} \\ (HA)^{⊗2} \ar[r]^{m̂_{2}} & HA}

The shift-graded linear map m̂_{2} can be obtained in a second way still. We will call the one resulting from the second construction m̃_{2} , with the aim of showing m̂_{2} =^{!} m̃_{2} .

Writing

Ψ_{2} := (q_{1} ⊗ q_{1}) · m_{2} ,

taking under consideration that m̃_{1} = 0, the Stasheff equation for morphisms the shiftgraded linear maps m̃_{2} and q_{2} are to satisfy reads

m̃_{2} · q_{1} − q_{2} · m_{1} =^{!} Ψ_{2} ;

cf. Example 22.(2).

We claim that Ψ_{2} factors over ι as Ψ_{2} = Ψ̌_{2} · ι. We have to show that Ψ_{2} · m_{1} = 0 ; cf. Problem 15.(1). In fact,

(q_{1} ⊗ q_{1}) · m_{2} · m_{1} =^{Ex. 21.(2)} (q_{1} ⊗ q_{1}) · (m_{1} ⊗ id) · m_{2} + (q_{1} ⊗ q_{1}) · (id ⊗m_{1}) · m_{2}
= (\underbrace{q_{1}m_{1}}_{= 0} ⊗q_{1}) · m_{2} + (q_{1} ⊗ \underbrace{q_{1}m_{1}}_{= 0}) · m_{2}
= 0 .

This proves the claim.

Letting

\begin{aligned}q_{2} & := & −Ψ̌_{2} · ν \\ m̃_{2} & := & Ψ̌_{2} · ρ , \end{aligned}

we obtain

m̃_{2} · q_{1} − q_{2} · m_{1} = Ψ̌_{2} · ρ · q_{1} + Ψ̌_{2} · ν · m_{1}
= Ψ̌_{2} · ρ · σ · ι + Ψ̌_{2} · (id −ρ · σ) · ι
= Ψ̌_{2} · ι
= Ψ_{2} ,

as required.

It remains to show

m̂_{2} =^{!} m̃_{2} .

It suffices to show that ρ^{⊗2}m̂_{2} =^{!} ρ^{⊗2}m̃_{2} , since ρ^{⊗2} is piecewise surjective; cf. Problem 16.(1). So we have to show that ρ^{⊗2}Ψ̌_{2}ρ =^{!} m̌_{2}ρ. It suffices to find a shift-graded linear map ξ : (ZA)^{⊗2} → A of degree −1 such that

ρ^{⊗2}Ψ̌_{2} − m̌_{2} =^{!} ξm_{1}|^{ZA} ,

i.e. such that

(ρ^{⊗2}Ψ̌_{2} − m̌_{2})ι =^{!} ξm_{1}|^{ZA}ι = ξm_{1} .

But

(ρ^{⊗2}Ψ̌_{2} − m̌_{2})ι
= ρ^{⊗2}Ψ̌_{2}ι − m̌_{2}ι
= ρ^{⊗2}Ψ_{2} − ι^{⊗2}m_{2}
= ρ^{⊗2}(q_{1} ⊗ q_{1})m_{2} − ι^{⊗2}m_{2}
= (ρ · σ · ι ⊗ ρ · σ · ι − ι ⊗ ι)m_{2}
= ((ι − νm_{1}) ⊗ (ι − νm_{1}) − ι ⊗ ι)m_{2}
= (−νm_{1} ⊗ ι − ι ⊗ νm_{1} + νm_{1} ⊗ νm_{1})m_{2}
= −(ν ⊗ ι)(m_{2}m_{1} − id ⊗m_{1}) − (ι ⊗ ν)(m_{2}m_{1} − (m_{1} ⊗ id)m_{2}) + (νm_{1} ⊗ ν)(m_{2}m_{1} − m_{1} ⊗ id)
= (−(ν ⊗ ι)m_{2} − (ι ⊗ ν)m_{2} + (νm_{1} ⊗ ν)m_{2})m_{1}

since ιm_{1} = 0. An associativity. We claim

(m̃_{2} ⊗ id − id ⊗m̃_{2})m̃_{2} =^{!} 0 .

It suffices to show that ρ^{⊗3}(m̃_{2} ⊗ id − id ⊗m̃_{2})m̃_{2} =^{!} 0, since ρ^{⊗3} is piecewise surjective; cf. Problem 16.(1). Now

ρ^{⊗3}(m̃_{2} ⊗ id − id ⊗m̃_{2})m̃_{2} = (ρ^{⊗2}m̃_{2} ⊗ ρ − ρ ⊗ ρ^{⊗2}m̃_{2})m̃_{2}
= (m̌_{2}ρ ⊗ ρ − ρ ⊗ m̌_{2}ρ)m̃_{2}
= (m̌_{2} ⊗ id − id ⊗m̌_{2})ρ^{⊗2}m̃_{2}
= (m̌_{2} ⊗ id − id ⊗m̌_{2})m̌_{2}ρ .

So it suffices to find a shift-graded linear map η : (ZA)^{⊗3} → A of degree −1 such that

(m̌_{2} ⊗ id − id ⊗m̌_{2})m̌_{2} =^{!} ηm_{1}|^{ZA}, i.e. such that (m̌_{2} ⊗ id − id ⊗m̌_{2})m̌_{2}ι =^{!} ηm_{1}|^{ZA}ι = ηm_{1}.

We obtain

(m̌_{2} ⊗ id − id ⊗m̌_{2})m̌_{2} · ι
= (m̌_{2} ⊗ id − id ⊗m̌_{2})ι^{⊗2}m_{2}
= (m̌_{2} · ι ⊗ ι − ι ⊗ m̌_{2} · ι)m_{2}
= (ι^{⊗2}m_{2} ⊗ ι − ι ⊗ ι^{⊗2}m_{2})m_{2}
= ι^{⊗3}(m_{2} ⊗ id − id ⊗m_{2})m_{2}
=^{Ex. 21.(3)} ι^{⊗3}(−(m_{1} ⊗ id^{⊗2}) · m_{3} − (id ⊗m_{1} ⊗ id) · m_{3} − (id^{⊗2} ⊗m_{1}) · m_{3} − m_{3} · m_{1})
= −ι^{⊗3} · m_{3} · m_{1}

since ιm_{1} = 0. Step 3. Write

Ψ_{3} := (−m̃_{2} ⊗ id + id ⊗m̃_{2})q_{2} + (q_{1} ⊗ q_{2}) · m_{2} − (q_{2} ⊗ q_{1}) · m_{2} + (q_{1} ⊗ q_{1} ⊗ q_{1}) · m_{3} .

We have to find m̃_{3} and q_{3} such that

m̃_{3} · q_{1} − q_{3} · m_{1} =^{!} Ψ_{3} ;

cf. Example 22.(3).

Provided we can show that Ψ_{3} ·m_{1} =^{!} 0, then we can write Ψ_{3} = Ψ̌_{3} ·ι; cf. Problem 15.(1). Then letting

\begin{aligned}q_{3} & := & −Ψ̌_{3} · ν \\ m̃_{3} & := & Ψ̌_{3} · ρ , \end{aligned}

we obtain

m̃_{3} · q_{1} − q_{3} · m_{1} = Ψ̌_{3} · ρ · q_{1} + Ψ̌_{3} · ν · m_{1}
= Ψ̌_{3} · ρ · σ · ι + Ψ̌_{3} · (id −ρ · σ) · ι
= Ψ̌_{3} · ρ · σ · ι + Ψ̌_{3} · (id −ρ · σ) · ι
= Ψ̌_{3} · ι
= Ψ_{3} ,

as required.

So it remains to show Ψ_{3} · m_{1} =^{!} 0. Plugging in q_{2} · m_{1} = m̃_{2} · q_{1} − (q_{1} ⊗ q_{1})m_{2} from Example 22.(2) and m_{2} · m_{1} = (m_{1} ⊗ id + id ⊗m_{1})m_{2} from Example 21.(2) and m_{3}m_{1} = −(m_{1} ⊗ id^{⊗2}) · m_{3} − (id ⊗m_{1} ⊗ id) · m_{3} − (id^{⊗2} ⊗m_{1}) · m_{3} − (m_{2} ⊗ id) · m_{2} + (id ⊗m_{2}) · m_{2} from Example 21.(3), using q_{1} ·m_{1} = σ ·ι·m_{1} = 0 as well as associativity of m̃_{2} , we obtain

Ψ_{3} · m_{1}

= −(m̃_{2} ⊗ id)q_{2} · m_{1} + (id ⊗m̃_{2})q_{2} · m_{1}

+(q_{1} ⊗ q_{2}) · m_{2} · m_{1} − (q_{2} ⊗ q_{1}) · m_{2} · m_{1} + (q_{1} ⊗ q_{1} ⊗ q_{1}) · m_{3} · m_{1}

= −(m̃_{2} ⊗ id)m̃_{2} · q_{1} + (m̃_{2} ⊗ id)(q_{1} ⊗ q_{1})m_{2}

+(id ⊗m̃_{2})m̃_{2} · q_{1} − (id ⊗m̃_{2})(q_{1} ⊗ q_{1})m_{2} +(q_{1} ⊗ q_{2})(m_{1} ⊗ id)m_{2} + (q_{1} ⊗ q_{2})(id ⊗m_{1})m_{2} −(q_{2} ⊗ q_{1})(m_{1} ⊗ id)m_{2} − (q_{2} ⊗ q_{1})(id ⊗m_{1})m_{2} −(q_{1} ⊗ q_{1} ⊗ q_{1})(m_{1} ⊗ id^{⊗2}) · m_{3} −(q_{1} ⊗ q_{1} ⊗ q_{1})(id ⊗m_{1} ⊗ id) · m_{3} −(q_{1} ⊗ q_{1} ⊗ q_{1})(id^{⊗2} ⊗m_{1}) · m_{3} −(q_{1} ⊗ q_{1} ⊗ q_{1})(m_{2} ⊗ id) · m_{2} +(q_{1} ⊗ q_{1} ⊗ q_{1})(id ⊗m_{2}) · m_{2}

= (m̃_{2} · q_{1} ⊗ q_{1})m_{2}

−(q_{1} ⊗ m̃_{2} · q_{1})m_{2} +(q_{1} ⊗ q_{2} · m_{1})m_{2} −(q_{2} · m_{1} ⊗ q_{1})m_{2} −((q_{1} ⊗ q_{1})m_{2} ⊗ q_{1})m_{2} +(q_{1} ⊗ (q_{1} ⊗ q_{1})m_{2}) · m_{2}

= ((m̃_{2} · q_{1} − q_{2} · m_{1} − (q_{1} ⊗ q_{1})m_{2}) ⊗ q_{1})m_{2}

(q_{1} ⊗ (−m̃_{2} · q_{1} + q_{2} · m_{1} + (q_{1} ⊗ q_{1})m_{2}))m_{2}

= 0 .

Note that the only nontrivial Stasheff equation for (m̃_{3}, m̃_{2}, m̃_{1}) takes place at k = 2, where it reads (m̃_{2} ⊗ id − id ⊗m̃_{2}) · m̃_{2} = 0, whose validity we have verified.

To directly proceed in this way, i.e. to construct Ψ_{n} analogously for n ⩾ 4 and to prove Ψ_{n} · m_{1} =^{!} 0 directly, seems to be involved. We will take a conceptual detour to prove the Theorem of Kadeishvili; cf. Theorem 50 below.

1.5 Getting rid of signs by conjugation

Let Z be a grading category.

Definition 35 Let A be a Z-graded module. Recall that A^{[1]} is the Z-graded module having

(A^{[1]})^{z} = A^{z[1]}

for z ∈ Mor(Z); cf. Definition 12. Define the shift-graded linear map

ω = ω_{A} : A^{[1]} → A

at z ∈ Mor(Z) by

\begin{matrix}(A^{[1]})^{z} & ⟶^{ω} & A^{z[1]} \\ a & ↦ & a . \end{matrix}

Lemma 36 Let n ∈ [0, ∞]. Let (A, (m_{ℓ})_{ℓ}) be a pre-A_{n}-algebra over Z. Given ℓ ∈ [1, n] ∩ Z, we write

ωm_{ℓ} := ω^{⊗ℓ} · m_{ℓ} · ω^{−} : (A^{[1]})^{⊗ℓ} → A^{[1]} ,

called the ω-conjugate of m_{ℓ}. Note that ^{ω}m_{ℓ} is of degree 1, independent of ℓ. Suppose given k ∈ [1, n] ∩ Z. Given (r, s, t) ⩾ (0, 1, 0) with r + s + t = k, we have

(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t} = ω^{⊗k}((−1)^{r+st}(id^{⊗r} ⊗m_{s} ⊗ id^{⊗t}) · m_{r+1+t})ω^{−} .

In particular, the Stasheff equation at k, viz.

0 = ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m_{s} ⊗ id^{⊗t}) · m_{r+1+t} ,

holds if and only if

0 = ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t}

holds.

Proof. Given (r, s, t) ⩾ (0, 1, 0) with r + s + t = k, we have

(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t}

= (id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ω^{⊗r+1+t} · m_{r+1+t} · ω^{−}
= (−1)^{r}(ω^{⊗r} ⊗ (ω^{⊗s} · m_{s}) ⊗ ω^{⊗t}) · m_{r+1+t} · ω^{−}
= ω^{⊗k}((−1)^{r+st}(id^{⊗r} ⊗m_{s} ⊗ id^{⊗t}) · m_{r+1+t})ω^{−} .

So

0 = ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t}

holds if and only if

0 = ω^{⊗k}( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m_{s} ⊗ id^{⊗t}) · m_{r+1+t})ω^{−}

holds, i.e. if and only if

0 = ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m_{s} ⊗ id^{⊗t}) · m_{r+1+t}

holds.

Lemma 37 Let n ∈ [0, ∞]. Let Ã = (Ã, (m̃_{ℓ})_{ℓ}) and A = (A, (m_{ℓ})_{ℓ}) be pre-A_{n}-algebras over Z. Let f = (f_{ℓ})_{ℓ} be a pre-A_{n}-morphism from Ã to A. Given ℓ ∈ [1, n] ∩ Z, we write

ωf_{ℓ} := ω^{⊗ℓ} · f_{ℓ} · ω^{−} : (Ã^{[1]})^{⊗ℓ} → A^{[1]} ,

called the ω-conjugate of f_{ℓ}. Note that ^{ω}f_{ℓ} is of degree 0, independent of ℓ. Suppose given k ∈ [1, n] ∩ Z. Given (r, s, t) ⩾ (0, 1, 0) with r + s + t = k, we have

(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t} = ω^{⊗k}((−1)^{r+st}(id^{⊗r} ⊗m̃_{s} ⊗ id^{⊗t}) · f_{r+1+t})ω^{−} .

Given r ∈ [1, k] and (i_{j})_{j∈[1,r]} ⩾ 0, we have

( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}) · ^{ω}m_{r} = ω^{⊗k}(⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋( ⨂_{j∈[1,r]} f_{i_{j}}) · m_{r})ω^{−} .

In particular, the Stasheff equation for morphisms at k, viz.

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m̃_{s} ⊗ id^{⊗t}) · f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋( ⨂_{j∈[1,r]} f_{i_{j}}) · m_{r}

holds if and only if

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}) · ^{ω}m_{r}

holds. Proof. Given (r, s, t) ⩾ (0, 1, 0) with r + s + t = k, we have

(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t}

= (id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ω^{⊗r+1+t} · f_{r+1+t} · ω^{−}
= (−1)^{r}(ω^{⊗r} ⊗ (ω^{⊗s} · m̃_{s}) ⊗ ω^{⊗t}) · f_{r+1+t} · ω^{−}
= ω^{⊗k}((−1)^{r+st}(id^{⊗r} ⊗m̃_{s} ⊗ id^{⊗t}) · f_{r+1+t})ω^{−} .

Given r ∈ [1, k] and (i_{j})_{j∈[1,r]} ⩾ 0, we have

( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}) · ^{ω}m_{r}

= ( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}) · ω^{⊗r} · m_{r} · ω^{−}
= ( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}ω) · m_{r} · ω^{−}
= ( ⨂_{j∈[1,r]} ω^{⊗i_{j}}f_{i_{j}}) · m_{r} · ω^{−}
= ω^{⊗k}(⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋( ⨂_{j∈[1,r]} f_{i_{j}}) · m_{r})ω^{−}

So

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ k=∑_{j} i_{j} = k \end{matrix}}( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}) · ^{ω}m_{r}

holds if and only if

ω^{⊗k}( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m̃_{s}⊗id^{⊗t})·f_{r+1+t})ω^{−} = ω^{⊗k}( ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}⌊(1−i_{j})_{j} , (i_{j})_{j}⌋( ⨂_{j∈[1,r]} f_{i_{j}})·m_{r})ω^{−}

holds, i.e. if and only if

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m̃_{s} ⊗ id^{⊗t}) · f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋( ⨂_{j∈[1,r]} f_{i_{j}}) · m_{r}

holds.

1.6 A tensor coalgebra interpretation

Let Z be a grading category.

Given a Z-graded module V and a, b ∈ Z_{⩾1} , we often abbreviate an elementary tensor as follows. Given v_{i} ∈ V for i ∈ [a, b], we write

v_{[a,b]}^{⊗} := v_{a} ⊗ v_{a+1} ⊗ … v_{b−1} ⊗ v_{b} .

Definition 38

(1) A coalgebra over Z is a Z-graded module T, equipped with a shift-graded linear map Δ : T → T ⊗ T of degree 0, called comultiplication, that is coassociative, i.e. that satisfies

Δ(Δ ⊗ id) = Δ(id ⊗Δ) .

\xymatrix{T ⊗ T \ar[r]^{id ⊗Δ} & T ⊗ T ⊗ T \\ T \ar[r]^{Δ} \ar[u]^{Δ} & T ⊗ T \ar[u]_{Δ⊗id}}

Often, we just write T = (T, Δ).

(2) Suppose given coalgebras T = (T, Δ) and T̃ = (T̃, Δ̃) over Z. A morphism of coalgebras, also called coalgebra morphism, (over Z) from T to T̃ is a shift-graded linear map ψ : T → T̃ of degree 0 such that

ψΔ̃ = Δ(ψ ⊗ ψ) .

\xymatrix{T ⊗ T \ar[r]^{ψ⊗ψ} & T ⊗ T \\ T \ar[r]^{ψ} \ar[u]^{Δ} & T \ar[u]_{Δ}}

(3) Suppose given a coalgebra T = (T, Δ) over Z. A coderivation on T is a shift-graded linear map T ⟶^{δ} T of degree 1 such that the co-Leibniz-rule

δΔ = Δ(id ⊗δ + δ ⊗ id)

holds.

Note that both sides are linear in δ, so that a linear combination of coderivations on T is again a coderivation on T.

A coderivation δ on T is called a codifferential if δ^{2} = 0.

(4) A coalgebra with codifferential over Z is a coalgebra T over Z, equipped with a codifferential δ on T. Often, we just write T = (T, Δ, δ).

(5) Suppose given coalgebras T = (T, Δ, δ) and T̃ = (T̃, Δ̃, δ̃) over Z. A morphism of

coalgebras with codifferential (over Z) is a coalgebra morphism T ⟶^{ψ} T̃ such that ψδ = δψ .

Lemma 39 (and Definition) Let V be a Z-graded module. Let n, ñ ∈ [0, ∞]. Consider the Z-graded module

T_{⩽n}(V) := ⨁_{k∈[1,n]∩Z} V^{⊗k} .

In particular, we often write

T(V) := T_{⩽∞}(V) = ⨁_{k∈Z_{⩾1}} V^{⊗k} .

Moreover, we identify

V = T_{⩽1}(V) .

(1) Let the shift-graded linear map Δ = Δ_{n,V} : T_{⩽n}(V) → T_{⩽n}(V) ⊗ T_{⩽n}(V) of degree 0

be defined at z ∈ Mor(Z) on the summand for k ∈ [1, n] ∩ Z, viz.

(V^{⊗k})^{z} = ⨁_{(y_{1},...,y_{k})∈fact_{k}(z)} ⨂_{i∈[1,k]} V^{y_{i}} ,

by defining it on its summand at (y_{1}, …, y_{k}) ∈ fact_{k}(z) by

\begin{matrix}⨂_{i∈[1,k]} V^{y_{i}} & → & (T_{⩽n}(V) ⊗ T_{⩽n}(V))^{z} \\ v_{[1,k]}^{⊗} = v_{1} ⊗ … ⊗ v_{k} & ↦ & ∑_{\begin{matrix}(i,j)∈Z_{⩾1}×Z_{⩾1} \\ i+j = k \end{matrix}} v_{1} ⊗ … ⊗ v_{i} ⊗ v_{i+1} ⊗ … ⊗ v_{i+j} = ∑_{\begin{matrix}(i,j)∈Z_{⩾1}×Z_{⩾1} \\ i+j = k \end{matrix}} v_{[1,i]}^{⊗} ⊗ v_{[i+1,i+j]}^{⊗} . \end{matrix}

Here the boldfaced tensor product symbol ⊗ merely indicates the summand the term is mapped to, that is v_{1} ⊗ … ⊗ v_{i} ⊗ v_{i+1} ⊗ … ⊗ v_{i+j} ∈ (V^{⊗i} ⊗ V^{⊗j})^{z}. Note that

v_{[1,k]}^{⊗}Δ ∈ (T_{⩽n−1}(V) ⊗ T_{⩽n−1}(V))^{z} .

So also the restricted shift-graded linear map Δ|^{T_{⩽n−1}(V)⊗T_{⩽n−1}(V)} of degree 0 exists. Then T_{⩽n}(V) = (T_{⩽n}(V), Δ_{n,V}) is a coalgebra, called the tensor coalgebra of V bounded by n. If n = ∞, we usually omit to mention that T(V) is bounded by ∞. Recall that for k ∈ [1, n] ∩ Z, we have shift-graded inclusion and projection maps V^{⊗k} ⟶^{ι_{k}} T_{⩽n}(V) ⟶^{π_{k}} V^{⊗k} of degree 0 ; cf. Problem 20.(1).

(2) Suppose given k ∈ [1, n] ∩ Z. The image of ι_{k}Δ : V^{⊗k} → T_{⩽n}(V) ⊗ T_{⩽n}(V) is contained in T_{⩽k−1}(V) ⊗ T_{⩽k−1}(V).

(3) Let the shift-graded linear map

T_{⩽n}(V) ⊗ T_{⩽ñ}(V) ⟶^{μ_{n,ñ,V}} T_{⩽n+ñ}(V)

of degree 0 be defined at z ∈ Mor(Z) for (u, ũ) ∈ fact_{2}(z), i.e. z = uũ, by

T_{⩽n}(V)^{u} ⊗ T_{⩽ñ}(V)ũ ⟶^{μ_{n,ñ,V}} T_{⩽n+ñ}(V)^{uũ} ,

which in turn on

(V^{⊗k})^{u} ⊗ (V^{⊗k̃})ũ ⟶^{μ_{n,ñ,V}} (V^{⊗k+k̃})^{uũ}

for k ∈ [1, n] and k̃ ∈ [1, ñ] is defined on the summand belonging to

(y_{1}, …, y_{k}) ∈ fact_{k}(u)
(ỹ_{1}, …, ỹk̃) ∈ factk̃(ũ)

by

\begin{matrix}(V^{y_{1}} ⊗ … ⊗ V^{y_{k}}) & ⊗ & (V ỹ_{1} ⊗ … ⊗ V ỹk̃) & ⟶^{μ_{n,ñ,V}} & V^{⊗k+k̃} \\ v_{[1,k]}^{⊗} & ⊗ & ṽ_{[1,k̃]}^{⊗} & ↦ & v_{[1,k]}^{⊗} ⊗ ṽ_{[1,k̃]}^{⊗} . \end{matrix}

(4) We have Kern Δ = V.

Proof. Ad (1). We have to show coassociativity of Δ. Let z ∈ Mor(Z). Let k ∈ [1, n] ∩ Z. Let (y_{1}, …, y_{k}) ∈ fact_{k}(z). Let v_{i} ∈ V^{y_{i}} for i ∈ [1, k]. Recall that we may abbreviate v_{[1,k]}^{⊗} = v_{1} ⊗ … v_{k}. On the one hand, we obtain

v_{[1,k]}^{⊗}Δ(id ⊗Δ) = ( ∑_{\begin{matrix}(i,j)⩾(1,1) \\ i+j = k \end{matrix}} v_{[1,i]}^{⊗} ⊗ v_{[i+1,i+j]}^{⊗})(id ⊗Δ)
= ∑_{\begin{matrix}(i,j)⩾(1,1) \\ i+j = k \end{matrix}} v_{[1,i]}^{⊗} ⊗ ( ∑_{\begin{matrix}(u,w)⩾(1,1) \\ u+w = j \end{matrix}} v_{[i+1,i+u]}^{⊗} ⊗ v_{[i+u+1,i+u+w]}^{⊗})
= ∑_{\begin{matrix}(i,u,w)⩾(1,1,1) \\ i+u+w = k \end{matrix}} v_{[1,i]}^{⊗} ⊗ v_{[i+1,i+u]}^{⊗} ⊗ v_{[i+u+1,i+u+w]}^{⊗} .

On the other hand, we obtain

v_{[1,k]}^{⊗}Δ(Δ ⊗ id) = ( ∑_{\begin{matrix}(i,j)⩾(1,1) \\ i+j = k \end{matrix}} v_{[1,i]}^{⊗} ⊗ v_{[i+1,i+j]}^{⊗})(Δ ⊗ id)

= ∑_{\begin{matrix}(i,j)⩾(1,1) \\ i+j = k \end{matrix}}( ∑_{\begin{matrix}(u,w)⩾(1,1) \\ u+w = i \end{matrix}} v_{[1,u]}^{⊗} ⊗ v_{[u+1,u+w]}^{⊗}) ⊗ v_{[i+1,i+j]}^{⊗}

= ∑_{\begin{matrix}(u,w,j)⩾(1,1,1) \\ u+w+j = k \end{matrix}} v_{[1,u]}^{⊗} ⊗ v_{[u+1,u+w]}^{⊗} ⊗ v_{[u+w+1,u+w+j]}^{⊗} .

So both results coincide. Hence Δ(id ⊗Δ) = Δ(Δ ⊗ id).

Ad (4). We have to show that (Kern Δ)^{z} =^{!} V^{z} for z ∈ Mor(Z).

Ad ⊇. Suppose given v_{1} ∈ V^{z}. Then v_{1}Δ = v_{[1,1]}^{⊗}Δ = ∑_{\begin{matrix}(i,j)⩾(1,1) \\ i+j = 1 \end{matrix}} v_{[1,i]}^{⊗} ⊗ v_{[i+1,i+j]}^{⊗} = 0 as an

empty sum.

Ad ⊆. Write Δ′ := Δ|^{T_{⩽n−1}(V)⊗T_{⩽n−1}(V)} ; cf. (1). We have to show that (Kern Δ′)^{z} ⊆^{!} V^{z}. Note that we have the shift-graded linear projection map T_{⩽n−1}(V) ⟶^{π_{1}} V^{⊗1} = V ; cf. Problem 20.(1). So we have

T_{⩽n}(V) ⟶^{Δ′} T_{⩽n−1}(V) ⊗ T_{⩽n−1}(V) ⟶^{π_{1}⊗id} V ⊗ T_{⩽n−1}(V) ⟶^{μ_{1,n−1,V}} V .

Let k ∈ [1, n] ∩ Z. Let (y_{1}, …, y_{k}) ∈ fact_{k}(z). Let v_{i} ∈ V^{y_{i}} for i ∈ [1, k]. If k ⩾ 2, then we obtain

v_{[1,k]}^{⊗}Δ′(π_{1} ⊗ id)μ_{1,n−1,V} = ( ∑_{\begin{matrix}(i,j)⩾(1,1) \\ i+j = k \end{matrix}} v_{[1,i]}^{⊗} ⊗ v_{[i+1,i+j]}^{⊗})(π_{1} ⊗ id)μ_{1,n−1,V}

= (v_{1} ⊗ v_{[2,k]}^{⊗})μ_{1,n−1,V}  (using k ⩾ 2)
= v_{[1,k]}^{⊗} .

So given ξ_{k} ∈ (V^{⊗k})^{z} for k ∈ [1, n] ∩ Z, with support { k ∈ [1, n] ∩ Z : k 0̸ } being finite, we let ξ := (ξ_{k})_{k∈[1,n]∩Z} and obtain

ξΔ′(π_{1} ⊗ id)μ_{1,n−1,V} = (ξ_{k})_{k∈[1,n]∩Z}Δ′(π_{1} ⊗ id)μ_{1,n−1,V} = (0) ⊔ (ξ_{k})_{k∈[2,n]∩Z} .

So if ξ ∈ (Kern Δ′)^{z}, we obtain ξ_{k} = 0 for k ∈ [2, n] ∩ Z, i.e. ξ ∈ V^{z}.

Corollary 41 Let n ∈ [1, ∞].

Suppose given a Z-graded module V.

Suppose given a Z-graded module U, an integer d ∈ Z and a shift-graded linear map U ⟶^{u} T_{⩽n}(V) of degree d.

Recall that we have shift-graded inclusion and projection maps V = V^{⊗1} ⟶^{ι_{1}} T_{⩽n}(V) ⟶^{π_{1}} V^{⊗1} = V of degree 0 ; cf. Problem 20.(1).

If u · Δ = 0, then u = u · π_{1} · ι_{1} .

Proof. By Lemma 39.(4), now V ⟶^{ι_{1}} T_{⩽n}(V) is the shift-graded linear inclusion map of the kernel in the sense of Problem 15.(1). Moreover, ι_{1} · π_{1} = id_{V} .

If uΔ = 0, then there exists a shift-graded linear map U ⟶^{ǔ} V of degree d such that u = ǔ · ι_{1} by loc. cit. Therefore u · π_{1} · ι_{1} = ǔ · ι_{1} · π_{1} · ι_{1} = ǔ · ι_{1} = u.

Proposition 42 (Lifting to coderivations)

Let n ∈ [1, ∞]. Let V be a Z-graded module.

Let

Coder_{n}(V) := { T_{⩽n}(V) ⟶^{δ} T_{⩽n}(V) : δ is a coderivation }
Coder_{n}^{red}(V) := { (V^{⊗k} ⟶^{μ_{k}} V)_{k∈[1,n]∩Z} : μ_{k} is a shift-graded linear map of degree 1 for k ∈ [1, n] }

So Coder_{n}(V) is a submodule of the module of all shift-graded linear map maps of degree 1 from T_{⩽n}(V) to T_{⩽n}(V). And Coder_{n}^{red}(V) is a module with linear combinations being formed entrywise.

We have the mutually inverse module morphisms

Coder_{n}(V)  ⟷^{∼}  Coder_{n}^{red}(V)
δ ⟼^{α = α_{Coder,n,V}} (ι_{k} · δ · π_{1})_{k∈[1,n]∩Z}
μβ ⟵^{β = β_{Coder,n,V}} μ = (μ_{k})_{k∈[1,n]∩Z} ,

where μβ is determined by

ι_{k} · (μβ) := ∑_{\begin{matrix}(r,s,t)∈Z_{⩾0}×Z_{⩾1}×Z_{⩾0} \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗μ_{s} ⊗ id^{⊗t}) · ι_{r+1+t} : V^{⊗k} → T_{⩽n}(V) .

for k ∈ [1, n] ∩ Z.

Proof.

Welldefinedness of β. Suppose given μ = (μ_{k})_{k} ∈ Coder_{n}^{red}(V). First, μβ is a shift-graded linear map of degree 1.

We need to show that μβ is a coderivation. Suppose given k ∈ [1, n] ∩ Z. Suppose given z ∈ Mor(Z) and (y_{1}, …, y_{k}) ∈ fact_{k}(z). Write y_{i} deg =: d_{i} for i ∈ [1, k]. Write d_{[a,b]} := ∑_{i∈[a,b]} d_{i} for a, b ∈ [1, k]. Suppose given v_{i} ∈ V^{y_{i}} for i ∈ [1, k]. We have to show that

v_{[1,k]}^{⊗}(μβ)Δ =^{!} v_{[1,k]}^{⊗}Δ(id ⊗(μβ) + (μβ) ⊗ id) .

In fact, we obtain

v_{[1,k]}^{⊗}(μβ)Δ

= v_{[1,k]}^{⊗}( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗μ_{s} ⊗ id^{⊗t}) · ι_{r+1+t})Δ
= ( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(−1)^{d_{[r+s+1,r+s+t]}} v_{[1,r]}^{⊗} ⊗ v_{[r+1,r+s]}^{⊗}μ_{s} ⊗ v_{[r+s+1,r+s+t]}^{⊗})Δ
= ∑_{\begin{matrix}(r′,r′′,s,t)⩾(1,0,1,0) \\ r′+r′′+s+t = k \end{matrix}}(−1)^{d_{[r′+r′′+s+1,r′+r′′+s+t]}} v_{[1,r′]}^{⊗} ⊗ v_{[r′+1,r′+r′′]}^{⊗} ⊗ v_{[r′+r′′+1,r′+r′′+s]}^{⊗}μ_{s} ⊗ v_{[r′+r′′+s+1,r′+r′′+s+t]}^{⊗}
+ ∑_{\begin{matrix}(r,s,t′,t′′)⩾(0,1,0,1) \\ r+s+t′+t′′ = k \end{matrix}}(−1)^{d_{[r+s+1,r+s+t′+t′′]}} v_{[1,r]}^{⊗} ⊗ v_{[r+1,r+s]}^{⊗}μ_{s} ⊗ v_{[r+s+1,r+s+t′]}^{⊗} ⊗ v_{[r+s+t′+1,r+s+t′+t′′]}^{⊗}
= ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = q \end{matrix}}(−1)^{d_{[p+r+s+1, p+r+s+t]}} v_{[1, p]}^{⊗} ⊗ v_{[p+1, p+r]}^{⊗} ⊗ v_{[p+r+1, p+r+s]}^{⊗}μ_{s} ⊗ v_{[p+r+s+1, p+r+s+t]}^{⊗}

+ ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = p \end{matrix}}(−1)^{d_{[r+s+1,r+s+t+q]}} v_{[1,r]}^{⊗} ⊗ v_{[r+1,r+s]}^{⊗}μ_{s} ⊗ v_{[r+s+1,r+s+t]}^{⊗} ⊗ v_{[r+s+t+1,r+s+t+q]}^{⊗})

= ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}(v_{[1, p]}^{⊗} ⊗ ( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = q \end{matrix}}(−1)^{d_{[p+r+s+1, p+r+s+t]}} v_{[p+1, p+r]}^{⊗} ⊗ v_{[p+r+1, p+r+s]}^{⊗}μ_{s} ⊗ v_{[p+r+s+1, p+r+s+t]}^{⊗})

+(−1)^{d_{[p+1,p+q]}}( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = p \end{matrix}}(−1)^{d_{[r+s+1,r+s+t]}} v_{[1,r]}^{⊗} ⊗ v_{[r+1,r+s]}^{⊗}μ_{s} ⊗ v_{[r+s+1,r+s+t]}^{⊗}) ⊗ v_{[p+1, p+q]}^{⊗})

= ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}(v_{[1, p]}^{⊗} ⊗ (v_{[p+1, p+q]}^{⊗}(μβ))

+(−1)^{d_{[p+1,p+q]}}(v_{[1, p]}^{⊗}(μβ)) ⊗ v_{[p+1, p+q]}^{⊗})

= ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}(v_{[1, p]}^{⊗} ⊗ v_{[p+1, p+q]}^{⊗})(id ⊗(μβ) + (μβ) ⊗ id)
= v_{[1,k]}^{⊗}Δ(id ⊗(μβ) + (μβ) ⊗ id) .

Composite β · α =^{!} id. Suppose given μ = (μ_{k})_{k} ∈ Coder_{n}^{red}(V). We obtain

μβα = (ι_{k} · (μβ) · π_{1})_{k}
= ( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗μ_{s} ⊗ id^{⊗t}) · ι_{r+1+t} · π_{1})_{k}
= (id^{⊗0} ⊗μ_{k} ⊗ id^{⊗0})_{k}
= μ .

Injectivity of α. Suppose given δ ∈ Coder_{n} V such that δα = 0. We have to show that

δ =^{!} 0.

By induction, we show that δ|_{T_{⩽ℓ}(V)} =^{!} 0 for ℓ ∈ [0, n] ∩ Z. Base of the induction. We have T_{⩽0}(V) = 0, whence δ|_{T_{⩽0}(V)} = 0. Step of the induction. Suppose given ℓ ∈ [0, n − 1] ∩ Z. Suppose that δ|_{T_{⩽ℓ}(V)} = 0. We have to show that ι_{ℓ+1} · δ =^{!} 0. Since ι_{ℓ+1}·Δ restricts to T_{⩽ℓ}(V)⊗T_{⩽ℓ}(V) in the target and since (id ⊗δ)|_{T_{⩽ℓ}(V)⊗T_{⩽ℓ}(V)} = 0 and (δ ⊗ id)|_{T_{⩽ℓ}(V)⊗T_{⩽ℓ}(V)} = 0, we have

ι_{ℓ+1} · δ · Δ = ι_{ℓ+1} · Δ · (id ⊗δ + δ ⊗ id) = 0 .

By Corollary 41, we conclude that

ι_{ℓ+1} · δ = ι_{ℓ+1} · δ · π_{1} · ι_{1} = 0 ,

the latter since δα = (ι_{k} · δ · π_{1})_{k} = 0. This concludes the induction. If n ∈ Z_{⩾1} , then letting ℓ = n, this shows δ = 0. If n = ∞, then ι_{ℓ} · δ = 0 for ℓ ∈ Z_{⩾1} , whence δ = 0. Proposition 43 (Lifting to coalgebra morphisms) Let n ∈ [1, ∞]. Let Ṽ and V be Z-graded modules. Let

Coalg_{n}(Ṽ , V) := { T_{⩽n}(Ṽ) ⟶^{ψ} T_{⩽n}(V) : ψ is a coalgebra morphism }
Coalg_{n}^{red}(Ṽ , V) := { (Ṽ^{⊗k} ⟶^{φ_{k}} V)_{k∈[1,n]∩Z} : φ_{k} is a shift-graded linear map of degree 0 for k ∈ [1, n] }

So Coalg_{n}(Ṽ , V) and Coalg_{n}^{red}(Ṽ , V) are sets. We have the mutually inverse bijections

Coalg_{n} V  ⟷^{∼}  Coalg_{n}^{red}
ψ ⟼^{α = α_{Coalg,n,Ṽ ,V}} (ι_{k} · ψ · π_{1})_{k∈[1,n]∩Z}
φβ ⟵^{β = β_{Coalg,n,Ṽ ,V}} φ = (φ_{k})_{k∈[1,n]∩Z} ,

where φβ is determined by

ι_{k} · (φβ) := ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}∈Z_{⩾1}^{×r} \\ ∑_{j∈[1,r]} i_{j} = k \end{matrix}}( ⨂_{j∈[1,r]} φ_{i_{j}}) · ι_{r} : Ṽ^{⊗k} → T_{⩽n}(V) .

for k ∈ [1, n] ∩ Z.

Proof.

Welldefinedness of β. Suppose given φ = (φ_{k})_{k} ∈ Coalg_{n}^{red}(Ṽ , V). First, φβ is a shiftgraded linear map of degree 0.

We need to show that φβ is a coalgebra morphism. Suppose given k ∈ [1, n] ∩ Z. Suppose given z ∈ Mor(Z) and (y_{1}, …, y_{k}) ∈ fact_{k}(z). Suppose given ṽ_{i} ∈ Ṽ^{y_{i}} for i ∈ [1, k]. We have to show that

ṽ_{[1,k]}^{⊗}(φβ)Δ =^{!} ṽ_{[1,k]}^{⊗}Δ((φβ) ⊗ (μβ)) .

Given r ∈ [1, k] and (i_{j})_{j∈[1,r]} ⩾ (1)_{j} such that ∑_{j} i_{j} = k and given s ∈ [1, r], we write

[i_{s}] := [ 1 + ∑_{j∈[1,s−1]} i_{j} , ∑_{j∈[1,s]} i_{j} ] .

We obtain

ṽ_{[1,k]}^{⊗}(φβ)Δ

= ṽ_{[1,k]}^{⊗}( ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}( ⨂_{j∈[1,r]} φ_{i_{j}}) · ι_{r})Δ

= ( ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}ṽ_{[i_{1}]}^{⊗}φ_{i_{1}} ⊗ … ⊗ ṽ_{[i_{r}]}^{⊗}φ_{i_{r}})Δ

= ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}} ∑_{\begin{matrix}(s,t)⩾(1,1) \\ s+t = r \end{matrix}}ṽ_{[i_{1}]}^{⊗}φ_{i_{1}} ⊗ … ⊗ ṽ_{[i_{s}]}^{⊗}φ_{i_{s}} ⊗ ṽ_{[i_{s+1}]}^{⊗}φ_{i_{s+1}} ⊗ … ⊗ ṽ_{[i_{s+t}]}^{⊗}φ_{i_{s+t}}

= ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}( ∑_{s∈[1,p]} ∑_{\begin{matrix}(i_{j})_{j∈[1,s]}⩾(1)_{j} \\ ∑_{j} i_{j} = p \end{matrix}}ṽ_{[i_{1}]}^{⊗}φ_{i_{1}} ⊗ … ⊗ ṽ_{[i_{s}]}^{⊗}φ_{i_{s}}) ⊗ ( ∑_{t∈[1,q]} ∑_{\begin{matrix}(i_{j})_{j∈[1,t]}⩾(1)_{j} \\ ∑_{j} i_{j} = q \end{matrix}}ṽ_{p+[i_{s+1}]}^{⊗}φ_{i_{s+1}} ⊗ … ⊗ ṽ_{p+[i_{s+t}]}^{⊗}φ_{i_{s+t}})

= ( ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}ṽ_{[1, p]}^{⊗}(φβ) ⊗ ṽ_{[p+1, p+q]}^{⊗}(φβ))

= ( ∑_{\begin{matrix}(p,q)⩾(1,1) \\ p+q = k \end{matrix}}ṽ_{[1, p]}^{⊗} ⊗ ṽ_{[p+1, p+q]}^{⊗})((φβ) ⊗ (φβ))

= ṽ_{[1,k]}^{⊗}Δ((φβ) ⊗ (μβ)) .

Composite β · α =^{!} id. Suppose given φ = (φ_{k})_{k} ∈ Coalg_{n}^{red}(Ṽ , V). We obtain

φβα = (ι_{k} · (φβ) · π_{1})_{k}
= ( ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}( ⨂_{j∈[1,r]} φ_{i_{j}}) · ι_{r} · π_{1})_{k}
= (⨂_{j∈[1,1]} φ_{k})_{k}
= φ .

Injectivity of α. Suppose given ψ, ψ′ ∈ Coalg_{n}(Ṽ , V) such that ψα = ψ′α. We have to show that ψ =^{!} ψ′. By induction, we show that ψ|_{T_{⩽ℓ}(Ṽ)} =^{!} ψ′|_{T_{⩽ℓ}(Ṽ)} for ℓ ∈ [0, n] ∩ Z. Base of the induction. We have T_{⩽0}(Ṽ) = 0, whence ψ|_{T_{⩽0}(Ṽ)} = 0 = ψ′|_{T_{⩽0}(Ṽ)} . Step of the induction. Suppose given ℓ ∈ [0, n−1]∩Z. Suppose that ψ|_{T_{⩽ℓ}(Ṽ)} = ψ′|_{T_{⩽ℓ}(Ṽ)} . We have to show that ι_{ℓ+1} · (ψ − ψ′) =^{!} 0. Note that ψ − ψ′ is only a shift-graded linear map of degree 0. We have

ι_{ℓ+1} · (ψ − ψ′) · Δ = ι_{ℓ+1} · ψ · Δ − ι_{ℓ+1} · ψ′ · Δ
= ι_{ℓ+1} · Δ · (ψ ⊗ ψ) − ι_{ℓ+1} · Δ · (ψ′ ⊗ ψ′)
= ι_{ℓ+1} · Δ · (ψ ⊗ (ψ − ψ′) + (ψ − ψ′) ⊗ ψ′)
= 0 ,

since ι_{ℓ+1} · Δ restricts in the target to T_{⩽ℓ}(Ṽ) ⊗ T_{⩽ℓ}(Ṽ) and since ψ − ψ′ vanishes on T_{⩽ℓ}(Ṽ). By Corollary 41, we conclude that

ι_{ℓ+1} · (ψ − ψ′) = ι_{ℓ+1} · (ψ − ψ′) · π_{1} · ι_{1} = 0 ,

the latter since (ι_{k} · δ · π_{1})_{k} = ψα = ψ′α = (ι_{k} · ψ′ · π_{1})_{k}. This concludes the induction. If n ∈ Z_{⩾1} , then letting ℓ = n, this shows ψ = ψ′. If n = ∞, then ι_{ℓ} · ψ = ι_{ℓ} · ψ′ for ℓ ∈ Z_{⩾1} , whence ψ = ψ′.

Corollary 44 Let n ∈ [1, ∞]. Let Ṽ and V be Z-graded modules. Suppose given k ∈ [1, n].

(1) Let δ : T_{⩽n}(V) → T_{⩽n}(V) be a coderivation. Then δ|_{T_{⩽k}(V)}^{T_{⩽k}(V)} exists.

(2) Let ψ : T_{⩽n}(Ṽ) → T_{⩽n}(V) be a coalgebra morphism. Then ψ|_{T_{⩽k}(Ṽ)}^{T_{⩽k}(V)} exists.

Proof. Ad (1). We have δ = δαβ, and

ι_{ℓ} · (δαβ) = ∑_{\begin{matrix}(r,s,t)∈Z_{⩾0}×Z_{⩾1}×Z_{⩾0} \\ r+s+t = ℓ \end{matrix}}(id^{⊗r} ⊗(ι_{s} · δ · π_{1}) ⊗ id^{⊗t}) · ι_{r+1+t}

maps to T_{⩽k}(V) for ℓ ∈ [1, k] ∩ Z ; cf. Proposition 42. Ad (2). We have ψ = ψαβ, and

ι_{ℓ} · (ψαβ) = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}∈Z_{⩾1}^{×r} \\ ∑_{j∈[1,r]} i_{j} = ℓ \end{matrix}}( ⨂_{j∈[1,r]}(ι_{i_{j}} · ψ · π_{1})) · ι_{r}

maps to T_{⩽k}(V) for ℓ ∈ [1, k] ∩ Z ; cf. Proposition 43.

Lemma 45 Let n ∈ [1, ∞]. Let Ṽ and V be Z-graded modules. Suppose given k ∈ [0, n − 1] ∩ Z.

(1) Suppose given a coderivation T_{⩽n}(V) ⟶^{δ} T_{⩽n}(V).

Suppose that δ^{2}|_{T_{⩽k}(V)} = 0.

Then ι_{k+1} · δ^{2} = ι_{k+1} · δ^{2} · π_{1} · ι_{1} .

(2) Suppose given coderivations T_{⩽n}(Ṽ) ⟶^{δ̃} T_{⩽n}(Ṽ) and T_{⩽n}(V) ⟶^{δ} T_{⩽n}(V). Suppose given a coalgebra morphism T_{⩽n}(Ṽ) ⟶^{ψ} T_{⩽n}(V). Suppose that (δ̃ · ψ − ψ · δ)|_{T_{⩽k}(V)} = 0. Then ι_{k+1} · (δ̃ · ψ − ψ · δ) = ι_{k+1} · (δ̃ · ψ − ψ · δ) · π_{1} · ι_{1} .

Proof. Ad (1). By Corollary 41, we need to show that ι_{k+1} · δ^{2} · Δ =^{!} 0. In fact, we get

ι_{k+1} · δ^{2} · Δ = ι_{k+1} · δ · δ · Δ
= ι_{k+1} · δ · Δ · (id ⊗δ + δ ⊗ id)
= ι_{k+1} · Δ · (id ⊗δ + δ ⊗ id) · (id ⊗δ + δ ⊗ id)
= ι_{k+1} · Δ · (id ⊗δ^{2} − δ ⊗ δ + δ ⊗ δ + δ^{2} ⊗ id)
= ι_{k+1} · Δ · (id ⊗δ^{2} + δ^{2} ⊗ id) ;

cf. Problem 6. Now ι_{k+1} · Δ restricts in the target to T_{⩽k}(V) ⊗ T_{⩽k}(V), so that we may conclude from δ^{2}|_{T_{⩽k}(V)} = 0 that ι_{k+1} · δ^{2} · Δ = 0. Ad (2). By Corollary 41, we need to show that ι_{k+1} · (δ̃· ψ − ψ · δ) · Δ =^{!} 0. In fact, we get

ι_{k+1} · (δ̃ · ψ − ψ · δ) · Δ
= ι_{k+1} · δ̃ · ψ · Δ − ι_{k+1} · ψ · δ · Δ
= ι_{k+1} · δ̃ · Δ · (ψ ⊗ ψ) − ι_{k+1} · ψ · Δ · (id ⊗δ + δ ⊗ id)
= ι_{k+1} · Δ · (id ⊗δ̃ + δ̃ ⊗ id) · (ψ ⊗ ψ) − ι_{k+1} · Δ · (ψ ⊗ ψ) · (id ⊗δ + δ ⊗ id)
= ι_{k+1} · Δ · (ψ ⊗ (δ̃ · ψ) + (δ̃ · ψ) ⊗ ψ − ψ ⊗ (ψ · δ) − (ψ · δ) ⊗ ψ)
= ι_{k+1} · Δ · (ψ ⊗ (δ̃ · ψ − ψ · δ) + (δ̃ · ψ − ψ · δ) ⊗ ψ .

Now ι_{k+1} · Δ restricts in the target to T_{⩽k}(V) ⊗ T_{⩽k}(V), so that we may conclude from (δ̃ · ψ − ψ · δ)|_{T_{⩽k}(V)} = 0 that ι_{k+1} · (δ̃ · ψ − ψ · δ) · Δ = 0.

Proposition 46 Let n ∈ [1, ∞]. Suppose given a pre-A_{n}-algebra (A, (m_{ℓ})_{ℓ}) over Z. Write

𝔪 := ((^{ω}m_{ℓ})_{ℓ})β_{Coder,n,A^{[1]}} ,

which is a coderivation on T_{⩽n}(A^{[1]}) ; cf. Proposition 42. The following assertions (1) and (2) are equivalent.

(1) The tuple (m_{ℓ})_{ℓ} satisfies the Stasheff equation at k ∈ [1, n]∩Z ; cf. Definition 19.(1).

(2) The coderivation 𝔪 is a codifferential, i.e. 𝔪^{2} = 0.

Proof. Suppose given u ∈ [0, n] ∩ Z. We claim equivalence of the following assertions (1_{u}) and (2_{u}).

(1_{u}) We have ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t} = 0 for k ∈ [1, u].

(2_{u}) We have 𝔪^{2}|_{T_{⩽u}(A^{[1]})} = 0.

We proceed by induction on u. For u = 0, both assertions (1_{0}) and (2_{0}) hold. Suppose given u ∈ [0, n − 1] ∩ Z. By induction, we suppose that the assertions (1_{u−1}) and (2_{u−1}) are equivalent. Consider the following assertions (i) and (ii).

(i) We have ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = u+1 \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t} = 0.

(ii) We have ι_{u+1} · 𝔪^{2} = 0.

We have to show that (2_{u}) ∧ (i) ⇔^{!} (2_{u}) ∧ (ii), for then

(1_{u+1}) ⇔ ((1_{u}) ∧ (i)) ⇔ ((2_{u}) ∧ (i)) ⇔ ((2_{u}) ∧ (ii)) ⇔ (2_{u+1}) .

We have

ι_{u+1} · 𝔪^{2} =^{L. 45.(1)} ι_{u+1} · 𝔪^{2} · π_{1} · ι_{1}
=^{P. 42}  ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ι_{r+1+t} · 𝔪 · π_{1} · ι_{1}
=^{P. 42}  ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t} · ι_{1} .

The needed equivalence now follows from ι_{1} being piecewise injective. This concludes the induction. This proves the claim. Case n ∈ Z. Letting u = n, the assertion of the Proposition follows by Lemma 36. Case n = ∞. We conclude as follows.

The tuple (m_{ℓ})_{ℓ} satisfies the Stasheff equation at k ∈ [1, ∞] ∩ Z .

⇔ The tuple (m_{ℓ})_{ℓ} satisfies the Stasheff equation at k ∈ [1, u] for u ∈ [0, ∞] ∩ Z.
⇔^{L. 36} We have ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t} = 0 for k ∈ [1, u].
⇔ We have 𝔪^{2}|_{T_{⩽u}(A^{[1]})} = 0 for u ∈ Z_{⩾0}.
⇔ We have ι_{ℓ} · 𝔪^{2} = 0 for ℓ ∈ Z_{⩾1}.
⇔ We have 𝔪^{2} = 0.

Proposition 47 Let n ∈ [1, ∞]. Suppose given pre-A_{n}-algebras Ã = (Ã, (m̃_{ℓ})_{ℓ}) and A = (A, (m_{ℓ})_{ℓ}) over Z. Suppose given a pre-A_{n}-morphism f = (f_{ℓ})_{ℓ} from Ã to A. Write

𝔪̃ := ((^{ω}m̃_{ℓ})_{ℓ})β_{Coder,n,Ã^{[1]}}
𝔪 := ((^{ω}m_{ℓ})_{ℓ})β_{Coder,n,A^{[1]}} ,

which are coderivations on T_{⩽n}(Ã^{[1]}) resp. on T_{⩽n}(A^{[1]}) ; cf. Proposition 42. Write

𝔣 := ((^{ω}f_{ℓ})_{ℓ})β_{Coalg,n,Ã^{[1]},A^{[1]}} ,

which is a coalgebra morphism from T_{⩽n}(Ã^{[1]}) to T_{⩽n}(A^{[1]}) ; cf. Proposition 43. The following assertions (1) and (2) are equivalent.

(1) The tuple (f_{ℓ})_{ℓ} satisfies the Stasheff equation for morphisms at k ∈ [1, n] ∩ Z ; cf. Definition 19.(1).

(2) The coalgebra morphism 𝔣 satisfies 𝔪̃ · 𝔣 = 𝔣 · 𝔪.

If Ã and A are A_{n}-algebras, (1) means that f is an A_{n}-morphism, whereas (2) means, using Proposition 46, that 𝔣 is a morphism of coalgebras with codifferential.

Proof. Suppose given u ∈ [0, n] ∩ Z. We claim equivalence of the following assertions (1_{u}) and (2_{u}).

(1_{u}) We have

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s}⊗id^{⊗t})·^{ω}f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{i∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}(^{ω}f_{i_{1}} ⊗…⊗^{ω}f_{i_{r}})·^{ω}m_{r+1+t}

for k ∈ [1, u].

(2_{u}) We have (𝔪̃ · 𝔣 − 𝔣 · 𝔪)|_{T_{⩽u}(Ã^{[1]})} = 0.

We proceed by induction on u. For u = 0, both assertions (1_{0}) and (2_{0}) hold. Suppose given u ∈ [0, n − 1] ∩ Z. By induction, we suppose that the assertions (1_{u−1}) and (2_{u−1}) are equivalent. Consider the following assertions (i) and (ii).

(i) We have

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = u+1 \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s}⊗id^{⊗t})·^{ω}f_{r+1+t} = ∑_{r∈[1,u+1]} ∑_{\begin{matrix}(i_{j})_{i∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = u+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗…⊗^{ω}f_{i_{r}})·^{ω}m_{r} .

(ii) We have ι_{u+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) = 0.

We have to show that (2_{u}) ∧ (i) ⇔^{!} (2_{u}) ∧ (ii), for then

(1_{u+1}) ⇔ ((1_{u}) ∧ (i)) ⇔ ((2_{u}) ∧ (i)) ⇔ ((2_{u}) ∧ (ii)) ⇔ (2_{u+1}) .

We have

ι_{u+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪)

=^{L. 45.(2)} ι_{u+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · π_{1} · ι_{1}
=^{P. 42, P. 43} ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ι_{r+1+t} · 𝔣 · π_{1} · ι_{1} − ∑_{r∈[1,u+1]} ∑_{\begin{matrix}(i_{j})_{i∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = u+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ι_{r} · 𝔪 · π_{1} · ι_{1}
=^{P. 42, P. 43} ( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t} − ∑_{r∈[1,u+1]} ∑_{\begin{matrix}(i_{j})_{i∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = u+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ^{ω}m_{r}) · ι_{1} .

The needed equivalence now follows from ι_{1} being piecewise injective. This concludes the induction. This proves the claim. Case n ∈ Z. Letting u = n, the assertion of the Proposition follows by Lemma 37. Case n = ∞. We conclude as follows.

The tuple (f_{ℓ})_{ℓ} satisfies the Stasheff equation for morphisms at k ∈ [1, ∞] ∩ Z

⇔ The tuple (f_{ℓ})_{ℓ} satisfies the Stasheff equation for morphisms at k ∈ [1, u] for u ∈ [0, ∞] ∩ Z
⇔^{L. 36} We have

∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = k \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t} = ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{i∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = k \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ^{ω}m_{r+1+t}

for k ∈ [1, u].

⇔ We have (𝔪̃ · 𝔣 − 𝔣 · 𝔪)|_{T_{⩽u}(Ã^{[1]})} = 0 for u ∈ Z_{⩾0}
⇔ We have ι_{ℓ} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) = 0 for ℓ ∈ Z_{⩾1}
⇔ We have 𝔪̃ · 𝔣 = 𝔣 · 𝔪 .

1.7 Kadeishvili’s theorem

With the coalgebra reinterpretation of §1.6 at hand, we can complete the task tentatively begun in Remark 30, which is to prove Kadeishvili’s theorem.

Let Z be a grading category.

Lemma 48 Let n ∈ [1, ∞]. Let Ã = (Ã, (m̃_{ℓ})_{ℓ}) be a pre-A_{n}-algebra over Z. Let A = (A, (m_{ℓ})_{ℓ}) be an A_{n}-algebra over Z. Let f = (f_{ℓ})_{ℓ} be a pre-A_{n}-morphisms from Ã to A that satisfies the Stasheff equation for morphisms at k ∈ [1, n] ∩ Z. Suppose that f_{1} is piecewise injective. Then Ã is an A_{n}-algebra. So then f : Ã → A is a morphism of A_{n}-algebras.

Proof. Using Propositions 42 and 43, we write

𝔪̃ := ((^{ω}m̃_{ℓ})_{ℓ})β_{Coder,n,Ã^{[1]}} (coderivation on T_{⩽n}(Ã^{[1]}) )
𝔪 := ((^{ω}m_{ℓ})_{ℓ})β_{Coder,n,A^{[1]}} (coderivation on T_{⩽n}(A^{[1]}) )
𝔣 := ((^{ω}f_{ℓ})_{ℓ})β_{Coalg,n,Ã^{[1]},A^{[1]}} (coalgebra morphism from T_{⩽n}(Ã^{[1]}) to T_{⩽n}(Ã^{[1]}) ) .

We have to show that 𝔪̃^{2} =^{!} 0 ; cf. Proposition 46. We claim that 𝔪̃^{2}|_{T_{⩽k}(Ã^{[1]})} =^{!} 0 for k ∈ [0, n] ∩ Z. We proceed by induction on k. For k = 0, we get T_{⩽0}(Ã^{[1]}), whence the assertion. Suppose given k ∈ [0, n − 1] ∩ Z. By induction, we have 𝔪̃^{2}|_{T_{⩽k}(Ã^{[1]})} = 0. We need to show that 𝔪̃^{2}|_{T_{⩽k+1}(Ã^{[1]})} =^{!} 0. It suffices to show that ι_{k+1} · 𝔪̃^{2} =^{!} 0. By Lemma 45.(1), we have ι_{k+1} · 𝔪̃^{2} · π_{1} · ι_{1} . Hence

0 =^{P. 46} ι_{k+1} · 𝔣 · 𝔪^{2} · π_{1}
=^{P. 47} ι_{k+1} · 𝔪̃ · 𝔣 · 𝔪 · π_{1}
=^{P. 47} ι_{k+1} · 𝔪̃^{2} · 𝔣 · π_{1}
=^{P. 47} ι_{k+1} · 𝔪̃^{2} · π_{1} · ι_{1} · 𝔣 · π_{1}
=^{P. 43} ι_{k+1} · 𝔪̃^{2} · π_{1} · ^{ω}f_{1} .

Since f_{1} is piecewise injective, so is ^{ω}f_{1}. Hence ι_{k+1} · 𝔪̃^{2} · π_{1} = 0. Thus

ι_{k+1} · 𝔪̃^{2} = ι_{k+1} · 𝔪̃^{2} · π_{1} · ι_{1} = 0 .

This proves the claim. If n ∈ Z_{⩾1} , then letting k = n, the claim gives 𝔪̃^{2} = 0. If n = ∞, then ι_{k} · 𝔪̃^{2} = 0 for k ∈ Z_{⩾1} , whence 𝔪̃^{2} = 0.

Lemma 49 Let n ∈ Z_{⩾1} .

Let Ã = (Ã, (m̃_{ℓ})_{ℓ∈[1,n+1]}) and A = (A, (m_{ℓ})_{ℓ∈[1,n+1]}) be pre-A_{n+1}-algebras over Z. Let f = (f_{ℓ})_{ℓ∈[1,n+1]} be a pre-A_{n+1}-morphism from Ã to A. Suppose that the following assertions (i, ii, iii) hold.

(i) (Ã, (m̃_{ℓ})_{ℓ∈[1,n]}) is an A_{n}-algebra, (A, (m_{ℓ})_{ℓ∈[1,n+1]}) is an A_{n+1}-algebra and (f_{ℓ})_{ℓ∈[1,n]} is an A_{n}-morphism from (Ã, (m̃_{ℓ})_{ℓ∈[1,n]}) to (A, (m_{ℓ})_{ℓ∈[1,n]}).

(ii) m̃_{1} = 0. (iii) f_{1} is a quasiisomorphism. Write

\begin{aligned}Ψ_{n+1} & := & − ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \\ (r,t)>(0,0) \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m̃_{s} ⊗ id^{⊗t}) · f_{r+1+t} \\ & & + ∑_{r∈[2,n+1]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n+1 \end{matrix}}⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋(f_{i_{1}} ⊗ … ⊗ f_{i_{r}}) · m_{r} . \end{aligned}

Then

Ψ_{n+1} · m_{1} = 0 .

Proof. Write

ωΨ_{n+1} := ω^{⊗n+1} · Ψ_{n} · ω^{−}
=^{L. 37} − ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \\ (r,t)>(0,0) \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t}

+ ∑_{r∈[2,n+1]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ^{ω}m_{r} .

Using Propositions 42 and 43, we write

𝔪̃ := ((^{ω}m̃_{ℓ})_{ℓ∈[1,n+1]})β_{Coder,n+1,Ã^{[1]}} (coderivation on T_{⩽n+1}(Ã^{[1]}) )
𝔪 := ((^{ω}m_{ℓ})_{ℓ∈[1,n+1]})β_{Coder,n+1,A^{[1]}} (coderivation on T_{⩽n+1}(A^{[1]}) )
𝔣 := ((^{ω}f_{ℓ})_{ℓ∈[1,n+1]})β_{Coalg,n+1,Ã^{[1]},A^{[1]}} (coalgebra morphism from T_{⩽n+1}(Ã^{[1]}) to T_{⩽n+1}(Ã^{[1]}) ) .

By Problem 22, we have

𝔪̃|_{T_{⩽n}(Ã^{[1]})}^{T_{⩽n}(Ã^{[1]})} := ((^{ω}m̃_{ℓ})_{ℓ∈[1,n]})β_{Coder,n+1,Ã^{[1]}} (coderivation on T_{⩽n}(Ã^{[1]}) )
𝔪|_{T_{⩽n}(A^{[1]})}^{T_{⩽n}(A^{[1]})} := ((^{ω}m_{ℓ})_{ℓ∈[1,n]})β_{Coder,n+1,A^{[1]}} (coderivation on T_{⩽n}(A^{[1]}) )
𝔣|_{T_{⩽n}(Ã^{[1]})}^{T_{⩽n}(A^{[1]})} := ((^{ω}f_{ℓ})_{ℓ∈[1,n]})β_{Coalg,n+1,Ã^{[1]},A^{[1]}} (coalgebra morphism from T_{⩽n}(Ã^{[1]}) to T_{⩽n}(Ã^{[1]}) ) .

So by (i), we have

\begin{aligned}𝔪̃^{2}|_{T_{⩽n}(Ã^{[1]})} & = & 0 \\ 𝔪^{2} & = & 0 \\ (𝔪̃ · 𝔣 − 𝔣 · 𝔪)|_{T_{⩽n}(Ã^{[1]})} & = & 0 ; \end{aligned}

cf. Propositions 46 and 47. Note that

ι_{n+1} · 𝔪̃ = ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = n+1 \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ι_{r+1+t}
=^{(ii)} ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ι_{r+1+t} .

In particular, (ι_{n+1} · 𝔪̃)|^{T_{⩽n}(Ã^{[1]})} exists. We obtain

ι_{n+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · π_{1}

=  ι_{n+1} · 𝔪̃ · 𝔣 · π_{1} − ι_{n+1}𝔣 · 𝔪 · π_{1}
=^{P. 42, P. 43} ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ι_{r+1+t} · 𝔣 · π_{1}

− ∑_{r∈[1,n+1]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ι_{r} · 𝔪 · π_{1}

=^{P. 42, P. 43} ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t}

− ∑_{r∈[1,n+1]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ^{ω}m_{r}

=  ^{ω}m̃_{n+1} · ^{ω}f_{1} + ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \\ (r,t)>0 \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t}

−^{ω}f_{n+1} · ^{ω}m_{1} − ∑_{r∈[2,n+1]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n+1 \end{matrix}}(^{ω}f_{i_{1}} ⊗ … ⊗ ^{ω}f_{i_{r}}) · ^{ω}m_{r}

=  −^{ω}Ψ_{n+1} + ^{ω}m̃_{n+1} · ^{ω}f_{1} − ^{ω}f_{n+1} · ^{ω}m_{1} .

Thus

−^{ω}Ψ_{n+1} · ^{ω}m_{1}
=  ι_{n+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · π_{1} · ^{ω}m_{1} − ^{ω}m̃_{n+1} · ^{ω}f_{1} · ^{ω}m_{1} + ^{ω}f_{n+1} · ^{ω}m_{1} · ^{ω}m_{1}
=^{(i)}  ι_{n+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · π_{1} · ^{ω}m_{1} − ^{ω}m̃_{n+1} · ^{ω}m̃_{1} ^{ω}f_{1} + ^{ω}f_{n+1} · 0
=^{(ii)}  ι_{n+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · π_{1} · ^{ω}m_{1}
=^{P. 42}  ι_{n+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · π_{1} · ι_{1} · 𝔪 · π_{1}
=^{L. 45.(2)}  ι_{n+1} · (𝔪̃ · 𝔣 − 𝔣 · 𝔪) · 𝔪 · π_{1}
=^{𝔪^{2} = 0}  ι_{n+1} · 𝔪̃ · 𝔣 · 𝔪 · π_{1}
=^{(ι_{n+1} · 𝔪̃)|^{T_{⩽n}(Ã^{[1]})}} ι_{n+1} · 𝔪̃ · 𝔪̃ · 𝔣 · π_{1}
=^{L. 45.(1)}  ι_{n+1} · 𝔪̃^{2} · π_{1} · ι_{1} · 𝔣 · π_{1}
=^{P. 43}  ι_{n+1} · 𝔪̃^{2} · π_{1} · ^{ω}f_{1} .

Therefore, we obtain

Ψ_{n+1} · m_{1} = −ω^{⊗n+1} · (^{ω}Ψ_{n+1} · ^{ω}m_{1}) · ω
= ω^{⊗n+1} · ι_{n+1} · 𝔪̃^{2} · π_{1} · ^{ω}f_{1} · ω
= (ω^{⊗n+1} · ι_{n+1} · 𝔪̃^{2} · π_{1} · ω) · f_{1} .

Hence it suffices to show the following claim. Claim. Suppose given Z-graded module T, an element d ∈ Z, a shift-graded linear map T ⟶^{ξ} A of degree d − 1 and a shift-graded linear map T ⟶^{η} Ã of degree d such that

ξ · m_{1} = η · f_{1} .

Then η = 0 and ξ · m_{1} = 0. It suffices to show η =^{!} 0. We use the notation of Remark 30. We have f_{1} · m_{1} = m̃_{1} · f_{1} =^{(ii)} 0. By Problem 15.(1), this yields the commutative diagram

[figure: Ã; f_{1}; A; ZÃ; f̌_{1}; ZA; ι; ρ; HÃ; Hf_{1}; ∼; HA .]

So we get the following commutative diagram.

[figure: T; η; ξ; Ã; f_{1}; m̃_{1} = 0; f̌_{1}; Hf_{1}; Ã; f_{1}; BA; ι̃; ZA; ι; ρ; HA; A; m_{1}; m_{1}|^{BA}; A; m_{1}; A]

We have

ξ · m_{1}|^{BA} · ι̃· ι = ξ · m_{1} = η · f_{1} = η · f̌_{1} · ι,

by pointwise injectivity of ι thus

ξ · m_{1}|^{BA} · ι̃ = η · f̌_{1} .

So

η · Hf_{1} = η · f̌_{1} · ρ = ξ · m_{1}|^{BA} · ι̃· ρ = 0 .

Since Hf_{1} is an isomorphism by (iii), we conclude that η = 0. This proves the claim.

In Lemma 49, it would have been sufficient to require Hf_{1} to be pointwise injective, for this suffices to prove the Claim.

Theorem 50 (Kadeishvili) Suppose that R is a field. Let n ∈ [1, ∞]. Recall that Z is a grading category. Let A = (A, (m_{ℓ})_{ℓ}) be an A_{n}-algebra over Z. There exist tuples of shift-graded linear maps (m̃_{ℓ})_{ℓ} and (q_{ℓ})_{ℓ} such that

HA = (HA, (m̃_{ℓ})_{ℓ})

is a minimal A_{n}-algebra over Z and such that

q := (q_{ℓ})_{ℓ} : HA → A

is a quasiisomorphism. Cf. Definitions 27.(6) and 29.(3). If A is unital and n ⩾ 2, then (HA, (m̃_{ℓ})_{ℓ}) and q = (q_{ℓ})_{ℓ} can be chosen to be unital; cf. Definitions 23 and 24.

When writing (m_{ℓ}^{HA})_{ℓ} instead of (m̃_{ℓ})_{ℓ}, no uniqueness is implied of this structure of an A_{n}-algebra on A with said properties.

Proof. We use the notation of Remark 30. Where necessary, we shall briefly recall arguments of Remark 31. First, we do not suppose A to be unital. We proceed by induction on n. Base. Suppose n = 1. Let m̃_{1} := 0. Let

q_{1} := σ · ι .

Note that q_{1} is piecewise injective since σ and ι are. Then m̃_{1} · q_{1} = 0 = σ · ι · m_{1} = q_{1} · m_{1} ; i.e. the Stasheff equation for morphisms holds at 1 ; cf. Example 22.(1). We have m̃_{1}^{2} = 0^{2} = 0 ; i.e. the Stasheff equation for HA holds at 1 ; cf. also Lemma 48. Since we have the commutative diagram

[figure: HA; q_{1}; A; ZHA; σ; ZA; ι; ρ; HHA; HA ,]

we have Hq_{1} = id_{HA}, which is an isomorphism. Step. Suppose the assertion to be known for n ∈ Z_{⩾1} . We have to show the assertion for n + 1. We have to show that there exists a shift-graded linear map q_{n+1} : (HA)^{⊗n+1} → A of degree −n and a shift-graded linear map m̃_{n+1} : (HA)^{⊗n+1} → HA of degree 1 − n such that (m̃_{ℓ})_{ℓ∈[1,n+1]} satisfies the Stasheff equation at n + 1 and such that (q_{ℓ})_{ℓ∈[1,n+1]} satisfies the Stasheff equation for morphisms at n + 1, with respect to (m̃_{ℓ})_{ℓ∈[1,n+1]} and (m_{ℓ})_{ℓ∈[1,n+1]} . Since q_{1} is piecewise injective, it suffices, by Lemma 48, to show the Stasheff equation for morphisms for (q_{ℓ})_{ℓ∈[1,n+1]} at n + 1. As in Lemma 49, we write

\begin{aligned}Ψ_{n+1} & := & − ∑_{\begin{matrix}(r,s,t)⩾(0,2,0) \\ r+s+t = n+1 \\ (r,t)>(0,0) \end{matrix}}(−1)^{r+st}(id^{⊗r} ⊗m̃_{s} ⊗ id^{⊗t}) · q_{r+1+t} \\ & & + ∑_{r∈[2,n+1]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n+1 \end{matrix}}⌊(1 − i_{j})_{j} , (i_{j})_{j}⌋(q_{i_{1}} ⊗ … ⊗ q_{i_{r}}) · m_{r} , \end{aligned}

which is a shift-graded linear map from (HA)^{⊗n+1} to A of degree 1 − n.

In this defining expression for Ψ_{n+1} , in fact m̃_{i} and q_{i} are involved only for i ∈ [1, n]. By Lemma 49, letting for the moment m̃_{n+1} and q_{n+1} be arbitrary, e.g. zero, we have

Ψ_{n+1} · m_{1} = 0 .

So

Ψ_{n+1} = Ψ̌_{n+1} · ι ,

where Ψ̌_{n+1} is a shift-graded linear map from (HA)^{⊗n+1} to ZA of degree 1 − n; cf. Problem 15.(1). Taking into account that m̃_{1} = 0, the Stasheff equation for morphisms at n + 1, which we have to show, writes

Ψ_{n+1} =^{!} m̃_{n+1} · q_{1} − q_{n+1} · m_{1} .

Let

\begin{aligned}q_{n+1} & := & −Ψ̌_{n+1} · ν \\ m̃_{n+1} & := & Ψ̌_{n+1} · ρ . \end{aligned}

Then

m̃_{n+1} · q_{1} − q_{n+1} · m_{1} = Ψ̌_{n+1} · ρ · σ · ι + Ψ̌_{n+1} · ν · m_{1}
=^{R. 30} Ψ̌_{n+1} · ρ · σ · ι + Ψ̌_{n+1} · (id_{ZA} −ρ · σ) · ι
= Ψ_{n+1} .

Second, we suppose A to be unital and n ⩾ 2. As in Remark 31, we obtain the following commutative diagram.

\xymatrix{A^{⊗2} \ar[r]^{m_{2}} & A \\ (ZA)^{⊗2} \ar[r]^{m̌_{2}} \ar[d]_{ρ^{⊗2}} \ar[u]^{ι^{⊗2}} & ZA \ar[d]^{ρ} \ar[u]_{ι} \\ (HA)^{⊗2} \ar[r]^{m̂_{2}} & HA}

Moreover, we get Ψ_{2} = (q_{1} ⊗ q_{1}) ⊗ m_{2} . Letting, as before, Ψ_{2} = Ψ̌_{2} · ι and q_{2} := −Ψ̌_{2} · ν and m̃_{2} := Ψ̌_{2} · ρ, we have

m̂_{2} = m̃_{2} ;

cf. Remark 31. For X ∈ Ob(Z), we have 1_{A,X} ∈ ZA; cf. Definition 23. So for X ⟶^{x} Y ⟶^{y} Z in Z, for a ∈ (ZA)^{x} and b ∈ (ZA)^{y}, we get

(1_{A,Y} ρ ⊗ bρ)m̃_{2} = (1_{A,Y} ⊗ b)ρ^{⊗2}m̃_{2} = (1_{A,Y} ⊗ b)m̌_{2}ρ = bρ ,

since

(1_{A,Y} ⊗ b)m̌_{2} = (1_{A,Y} ⊗ b)m̌_{2}ι = (1_{A,Y} ⊗ b)ι^{⊗2}m_{2} = (1_{A,Y} ⊗ b)m_{2} = b .

Likewise, we get

(aρ ⊗ 1_{A,Y} ρ)m̃_{2} = aρ .

So the element 1_{A,Y} ρ ∈ (HA)^{id^{Y}} is neutral, i.e. 1_{HA,Y} = 1_{A,Y} ρ. Hence the A_{n}-algebra HA = (HA, (m̃_{ℓ})_{ℓ}) is unital. By Problem 18, the choice of σ made in Remark 30 can be made in such a way that 1_{HA,X}σ = 1_{A,X}ρσ = 1_{A,X} for X ∈ Ob(Z), whence

1_{HA,X}q_{1} = 1_{HA,X}σ ι = 1_{A,X}ι = 1_{A,X} .

Therefore the A_{n}-morphism q = (q_{ℓ})_{ℓ} is unital.

In the induction step of Theorem 50, we could have used an arbitrary shift-graded linear map Ψ_{n+1} from (HA)^{⊗n+1} to A of degree 1 − n that satisfies Ψ_{n+1} · m_{1} = 0 and define q_{n+1} := −Ψ̌_{n+1} · ν and m̃_{n+1} := Ψ̌_{n+1} · ρ. Lemma 49 merely guarantees the existence of such a shift-graded linear map.

Remark 51 Let G be a finite group. Let N ∈ Z_{⩾1} . Let M_{1} , …, M_{N} be RG-modules. Suppose M_{1} = R to carry the trivial RG-module structure, i.e. gr = r for g ∈ G and r ∈ R = M_{1} . Let P_{s} be a projective resolution of M_{s} for s ∈ [1, N]. Write \underline{P} := (P_{s})_{s∈[1,N]} . Let A := Hom_{RG}(\underline{P}) be the regular differential graded category of \underline{P} ; cf. Lemma 28. So A is a unital A_{∞}-algebra over

Z := Z × [1, N]^{×2} .

For (j, (s, t)) ∈ Mor(Z), we get

(HA)^{j,(s,t)} = _{K}( P_{s} , P_{t}^{[j]} ) =: Ext_{RG}^{j}(M_{s}, M_{t}) ,

where we have written K := K(RG -Mod) ; cf. Problem 14.(2). In particular,

(HA)^{j,(1,1)} = _{K}( P_{1} , P_{1}^{[j]} ) =: Ext_{RG}^{j}(M_{1}, M_{1}) = Ext_{RG}^{j}(R, R) = H^{j}(G; R) ,

the group cohomology of G over the ground ring R. Now suppose R to be a field. Kadeishvili’s Theorem 50 yields the structure (m̃_{ℓ})_{ℓ} of a minimal A_{∞}-algebra over Z on HA and a unital quasiisomorphism

HA → A .

In particular, m̃_{1} = 0. Moreover,

Ext_{RG}^{j}(M_{s}, M_{t}) ⊗ Ext_{RG}^{k}(M_{t}, M_{u}) ⟶^{m̃_{2}} Ext_{RG}^{j+k}(M_{s}, M_{u})
[f] ⊗ [g]  ↦ [f · g^{[j]}]

is the Yoneda product, where (j, (s, t)), (k, (t, u)) ∈ Mor(Z). Cf. Lemma 28, Remark 31. In particular,

H^{j}(G; R) ⊗ H^{j}(G; R) ⟶^{m̃_{2}} H^{j}(G; R)

is also known as cup product. In that sense, m̃_{n} for n ∈ Z_{⩾3} are sometimes referred to as “higher” cup products on the cohomology ring of G over the ground field R.

Chapter 2

Schmid’s extension of Kadeishvili

The purpose of the extra machinery in this §2.1 is to remove the restriction on R to be a field from Theorem 50.

Let Z be a grading category.

2.1 Split-filtered Z-graded modules

Definition 52 A split-filtered Z-graded module is a Z-graded module M, together with a tuple (M^{⟨i⟩})_{i∈Z} of Z-graded submodules of M such that the following conditions (1, 2) hold.

(1) We have M^{⟨i⟩} = 0 for i ∈ Z_{<0}.

(2) We have M = ⨁_{i∈Z_{⩾0}} M^{⟨i⟩}.

We often abbreviate M = (M, (M^{⟨i⟩})_{i}). Write M^{⩽k} := ⨁_{i∈[0,k]} M^{⟨i⟩} for k ∈ Z. So M^{⩽k} is a Z-graded submodule of M. We have shift-graded linear inclusion and projection maps

M^{⟨i⟩} ⟶^{ι_{M}^{⟨i⟩}} M ⟶^{π_{M}^{⟨i⟩}} M^{⟨i⟩}

of degree 0 for i ∈ Z. We often abbreviate ι^{⟨i⟩} = ι_{M}^{⟨i⟩} and π^{⟨i⟩} = π_{M}^{⟨i⟩}. So ι^{⟨i⟩}π^{⟨i⟩} = id_{M^{⟨i⟩}} for i ∈ Z and ι^{⟨i⟩}π^{⟨j⟩} = 0 for i, j ∈ Z with i ≠ j. With a similar abuse of notation, we also have shift-graded linear inclusion and projection maps

M^{⟨i⟩} ⟶^{ι^{⟨i⟩}} M^{⩽k} ⟶^{π^{⟨i⟩}} M^{⟨i⟩}

of degree 0 for i ∈ [0, k].

Given k, ℓ ∈ Z such that ℓ ⩽ k, we also have the shift-graded linear inclusion and projection maps

M^{⩽ℓ} ⟶^{ι^{⩽ℓ}} M^{⩽k} ⟶^{π^{⩽ℓ}} M^{⩽ℓ}

of degree 0.

I do not know whether a variant of the theory can be carried through with filtered Z-graded modules instead of split-filtered Z-graded modules.

Example 53 Let X be a Z-graded module. For z ∈ Mor(Z), choose

… → P^{⟨2⟩,z[−2]} → P^{⟨1⟩,z[−1]} → P^{⟨0⟩,z[0]} → X^{z} → 0

to be an augmented projective resolution of X^{z} (over R), i.e. P^{⟨k⟩,z[−k]} is projective for k ∈ Z_{⩾0} and the sequence is exact at each position.

Write P^{z} := ⨁_{i∈Z_{⩾0}} P^{⟨i⟩,z} for z ∈ Mor(Z).

Then P is a split-filtered Z-graded module with P^{⟨i⟩} := (P^{z,⟨i⟩})_{z∈Mor(Z)} for i ∈ Z.

If Z = Z, we can picture the components of P as follows.

⋮  ⋮  ⋮ … P^{⟨2⟩,1} P^{⟨1⟩,1} P^{⟨0⟩,1}  this row sums up to P^{1} … P^{⟨2⟩,0} P^{⟨1⟩,0} P^{⟨0⟩,0}  this row sums up to P^{0} … P^{⟨2⟩,−1} P^{⟨1⟩,−1} P^{⟨0⟩,−1}  this row sums up to P^{−1} … P^{⟨2⟩,−2} P^{⟨1⟩,−2} P^{⟨0⟩,−2}  this row sums up to P^{−2} ⋮  ⋮  ⋮

[figure]

Note that the objects of the respectively chosen projective resolutions can be found in the diagonals of this diagram, such as the boxed one, whose objects belong to a projective resolution of X^{1}.

2.2 eA_{∞}-algebras and eA_{∞}-categories

Definition 54 Let n ∈ [0, ∞]. An eA_{n}-algebra over Z is a split-filtered Z-graded module A = (A, (A^{⟨i⟩})_{i}), together with the structure of an A_{n}-algebra (m_{ℓ})_{ℓ} on the Z-graded module A, such that the Schmid condition

( ⨂_{j∈[1,k]} A^{⟨i_{j}⟩})m_{k} ⊆ A^{⩽2k−2+∑_{j∈[1,k]} i_{j}}

holds for k ∈ [1, n] ∩ Z and (i_{j})_{j∈[1,k]} ∈ Z_{⩾0}^{×k}. We often abbreviate A = (A, (m_{ℓ})_{ℓ∈[1,n]}, (A^{⟨i⟩})_{i∈Z}). We often write (A^{⟨i⟩})^{z} =: A^{⟨i⟩,z} for i ∈ Z and z ∈ Mor(Z). We often write (A^{⩽k})^{z} =: A^{⩽k,z} for k ∈ Z and z ∈ Mor(Z).

The “e” in “eA_{n}-algebra” stands for “extended”.

Schmid states that the Schmid condition was motivated by Sagave; cf. [5, Def. 76, (EA 3)], [3, Def. 2.1]. It is a bit weaker than Sagave’s implicitly stated condition.

Definition 55 Let n ∈ [0, ∞]. An eA_{n}-algebra A = (A, (m_{ℓ})_{ℓ∈[1,n]}, (A^{⟨i⟩})_{i∈Z}) over Z is called minimal if the strong Schmid condition

( ⨂_{j∈[1,k]} A^{⟨i_{j}⟩})m_{k} ⊆ A^{⩽2k−3+∑_{j∈[1,k]} i_{j}}

holds for k ∈ [1, n] ∩ Z and (i_{j})_{j∈[1,k]} ∈ Z_{⩾0}^{×k}.

Remark 56 Let n ∈ [1, ∞]. Let A = (A, (m_{ℓ})_{ℓ∈[1,n]}, (A^{⟨i⟩})_{i∈Z}) be an eA_{n}-algebra.

(1) For k = 1, the Schmid condition reads A^{⟨i⟩}m_{1} ⊆ A^{⩽i} for i ∈ Z_{⩾0} .

So if Z = Z, taking into account that m_{1} is of degree 1, the (possibly nonvanishing) components of m_{1} can be visualised as follows.

[figure: ⋮; ⋮; ⋮; …; A^{⟨2⟩,1}; A^{⟨1⟩,1}; A^{⟨0⟩,1}; …; A^{⟨2⟩,0}; A^{⟨1⟩,0}; A^{⟨0⟩,0}; …; A^{⟨2⟩,−1}; A^{⟨1⟩,−1}; A^{⟨0⟩,−1}; …; A^{⟨2⟩,−2}; A^{⟨1⟩,−2}; A^{⟨0⟩,−2}; ⋮; ⋮; ⋮]

By this, we mean that ι^{⟨i⟩}m_{1}π^{⟨j⟩} = 0 unless j ∈ [0, i]. E.g. on A^{⟨2⟩}, the shift-graded linear map m_{1} of degree 1 has the components

ι^{⟨2⟩} · m_{1} · π^{⟨2⟩} , ι^{⟨2⟩} · m_{1} · π^{⟨1⟩} , ι^{⟨2⟩} · m_{1} · π^{⟨0⟩} ,

all others vanish. The strong Schmid condition reads A^{⟨i⟩}m_{1} ⊆ A^{⩽i−1} for i ∈ Z_{⩾0} . So in case A is minimal, the components of m_{1} can be visualised as follows.

\xymatrix{ & ⋮ & ⋮ & ⋮ \\ … & A^{⟨2⟩,1} & A^{⟨1⟩,1} & A^{⟨0⟩,1} \\ … & A^{⟨2⟩,0} \ar[ur] \ar[urr] & A^{⟨1⟩,0} \ar[ur] & A^{⟨0⟩,0} \\ … & A^{⟨2⟩,−1} \ar[ur] \ar[urr] & A^{⟨1⟩,−1} \ar[ur] & A^{⟨0⟩,−1} \\ … & A^{⟨2⟩,−2} \ar[ur] \ar[urr] & A^{⟨1⟩,−2} \ar[ur] & A^{⟨0⟩,−2} \\  & ⋮ & ⋮ & ⋮}

(2) Suppose that n ⩾ 2. For k = 2, the Schmid condition reads

(A^{⟨i_{1}⟩} ⊗ A^{⟨i_{2}⟩})m_{2} ⊆ A^{⩽2+i_{1}+i_{2}}

for i_{1} , i_{2} ∈ Z_{⩾0} .

In case of A being minimal, the strong Schmid condition reads

(A^{⟨i_{1}⟩} ⊗ A^{⟨i_{2}⟩})m_{2} ⊆ A^{⩽1+i_{1}+i_{2}}

for i_{1} , i_{2} ∈ Z_{⩾0} .

Remark 57 Let n ∈ [0, ∞]. Suppose given an A_{n}-algebra A′ = (A′, (m′_{ℓ})_{ℓ}). Define an eA_{n}-algebra A = (A, (m_{ℓ})_{ℓ}, (A^{⟨i⟩})_{i}) by letting A = A′ as Z-graded modules, by letting m_{ℓ} := m′_{ℓ} for ℓ ∈ Z_{⩾1} and by letting

A^{⟨i⟩} := \{ \begin{matrix}A′ & if i = 0 \\ 0 & if i ∈ Z ∖ \{0\} \end{matrix}

for i ∈ Z. In fact, we have ⨁_{i∈Z} A^{⟨i⟩} = A^{⟨0⟩} = A′ = A. We have to verify the Schmid condition. For k ∈ [1, n] ∩ Z and i_{1} , …, i_{k} ∈ Z_{⩾0} , we obtain

(A^{⟨i_{1}⟩} ⊗ … ⊗ A^{⟨i_{k}⟩})m_{k} \{ \begin{matrix}= & 0 & if there exists j ∈ [1, k] with i_{j} ⩾ 1 \\ ⊆ & A^{⟨0⟩} = A^{⩽2k−2+∑_{j} i_{j}} & if i_{1} = ⋯ = i_{k} = 0 , \end{matrix}

since in the second case, we have 2k − 2 ⩾ 0 and ∑_{j} i_{j} = 0. Now the eA_{n}-algebra A is minimal if and only if (A^{⟨0⟩} ⊗ … ⊗ A^{⟨0⟩})m_{k} ⊆ A^{⩽2k−3} for k ∈ Z_{⩾1}. Since 2k − 3 ⩾ 0 and thus A^{⩽2k−3} = A^{⟨0⟩} if k ⩾ 2, this condition is equivalent to A^{⟨0⟩}m_{1} ⊆ A^{⩽−1} = 0, i.e. to A′m′_{1} = 0, i.e. to A′ being a minimal A_{n}-algebra. For short,

A minimal ⇔ A′ minimal

2.3 A base of an induction

Remark 58 (and definition) Let A = (A, (m_{1}), (A^{⟨i⟩})_{i}) be a minimal eA_{1}-algebra. We have (A^{⟨i⟩})m_{1} ⊆ A^{⩽i−1} for i ∈ Z_{⩾0} ; cf. Remark 56.(1). That is, we have

ι^{⟨i⟩} · m_{1} = ∑_{j∈[0,i−1]} ι^{⟨i⟩} · m_{1} · π^{⟨j⟩} · ι^{⟨j⟩}

for i ∈ Z_{⩾0} . For i ∈ Z_{⩾}0, we consider the shift-graded linear map

m_{1}^{⟨i⟩} := ι^{⟨i⟩} · m_{1} · π^{⟨i−1⟩} : A^{⟨i⟩} → A^{⟨i−1⟩}

of degree 1. For i ∈ Z_{⩾1}, we have

m_{1}^{⟨i⟩} · m_{1}^{⟨i−1⟩} = ι^{⟨i⟩} · m_{1} · π^{⟨i−1⟩} · ι^{⟨i−1⟩} · m_{1} · π^{⟨i−2⟩}
= ι^{⟨i⟩} · \overbrace{m_{1} · m_{1}}^{= 0 by Stasheff} ·π^{⟨i−2⟩}

− ∑_{j∈[0,i−2]} ι^{⟨i⟩} · m_{1} · π^{⟨j⟩} · \underbrace{ι^{⟨j⟩} · m_{1} · π^{⟨i−2⟩}}_{= 0 since j ⩽ i − 2}

= 0 .

If Im(m_{1}^{⟨i⟩}) = Kern(m_{1}^{⟨i−1⟩}) for i ∈ Z_{⩾2} , then A is called diagonally resolving. Cf. Example 53.

Lemma 59 Let A = (A, (m_{1}), (A^{⟨i⟩})_{i}) be a diagonally resolving minimal eA_{1}-algebra.

(1) We have ZA = A^{⟨0⟩} + BA.

(2) We have A^{⩽k}m_{1} = BA ∩ A^{⩽k−1} for k ∈ Z.

Proof. Ad (1). Ad ⊇. We have A^{⟨0⟩}m_{1} ⊆ A^{⩽−1} = 0, whence A^{⟨0⟩} ⊆ ZA. So A^{⟨0⟩} + BA ⊆ ZA. Ad ⊆. We claim that (A^{⩽j} ∩ ZA) + BA ⊆^{!} (A^{⩽j−1} ∩ ZA) + BA for j ∈ Z_{⩾1} . Suppose given z ∈ Mor(Z). Suppose given a ∈ (A^{⩽j} ∩ ZA)^{z} = A^{⩽j,z} ∩ (ZA)^{z}. We have to show that a ∈^{!} ((A^{⩽j−1} ∩ ZA) + BA)^{z} = (A^{⩽j−1,z} ∩ (ZA)^{z}) + (BA)^{z}. Since a ∈ (ZA)^{z} and since (BA)^{z} ⊆ (ZA)^{z}, it suffices to show that a ∈^{!} A^{⩽j−1,z} + (BA)^{z}. We have

(aπ^{⟨j⟩})m_{1}^{⟨j⟩} = aπ^{⟨j⟩}ι^{⟨j⟩}m_{1}π^{⟨j−1⟩}
= (a − ∑_{i∈[0,j−1]} aπ^{⟨i⟩}ι^{⟨i⟩})m_{1}π^{⟨j−1⟩}
= − ∑_{i∈[0,j−1]} aπ^{⟨i⟩}ι^{⟨i⟩}m_{1}π^{⟨j−1⟩}
= 0 ,

since i ⩽ j −1 for i ∈ [0, j −1], i.e. aπ^{⟨j⟩}ι^{⟨j⟩} ∈ Kern(m_{1}^{⟨j⟩})^{z}. Since A is diagonally resolving and since j ⩾ 1, we conclude that aπ^{⟨j⟩} ∈ Im(m_{1}^{⟨j+1⟩})^{z[−1]}. So there exists a′ ∈ A^{⟨j+1⟩,z[−1]} such that

aπ^{⟨j⟩} = a′m_{1}^{⟨j+1⟩} = a′ι^{⟨j+1⟩}m_{1}π^{⟨j⟩} .

Now

a = aπ^{⟨j⟩}ι^{⟨j⟩} + ( ∑_{i∈[0,j−1]} aπ^{⟨i⟩}ι^{⟨i⟩})
= a′ι^{⟨j+1⟩}m_{1}π^{⟨j⟩}ι^{⟨j⟩} + ( ∑_{i∈[0,j−1]} aπ^{⟨i⟩}ι^{⟨i⟩})
= \underbrace{a′ι^{⟨j+1⟩}m_{1}}_{∈(BA)^{z}} − \underbrace{( ∑_{i∈[0,j−1]} a′ι^{⟨j+1⟩}m_{1}π^{⟨i⟩}ι^{⟨i⟩})}_{∈A^{⩽j−1,z}} + \underbrace{( ∑_{i∈[0,j−1]} aπ^{⟨i⟩}ι^{⟨i⟩})}_{∈A^{⩽j−1,z}} .

This proves the claim. Given z ∈ Mor(Z) and a ∈ (ZA)^{z}, we have to show that z ∈^{!} (A^{⟨0⟩}+BA)^{z} = A^{⟨0⟩,z} +(BA)^{z}. There exists j ∈ Z_{⩾1} such that a ∈ A^{⩽j,z}. So

a ∈ (A^{⩽j,z} ∩ (ZA)^{z}) + (BA)^{z}
⊆^{Claim} (A^{⩽j−1,z} ∩ (ZA)^{z}) + (BA)^{z}
⊆^{Claim} …
⊆^{Claim} (A^{⩽0,z} ∩ (ZA)^{z}) + (BA)^{z}
= A^{⟨0⟩,z} + (BA)^{z} .

Ad (2). Ad ⊆. We have A^{⩽k}m_{1} ⊆ BA. We have A^{⩽k}m_{1} ⊆ A^{⩽k−1} by minimality of A.

Ad ⊇. Claim. Given j ∈ Z_{⩾0} and ℓ ∈ Z_{⩾j+1} , we have A^{⩽ℓ}m_{1} ∩A^{⩽j−1} ⊆^{!} A^{⩽ℓ−1}m_{1} ∩A^{⩽j−1}.

Suppose given z ∈ Mor(Z) and a ∈ A^{⩽ℓ,z[−1]} such that am_{1} ∈ A^{⩽j−1,z}. We have to show

that am_{1} ∈^{!} A^{⩽ℓ−1,z[−1]}m_{1} .

Note that am_{1} ∈ A^{⩽j−1,z} ⊆ A^{⩽ℓ−2,z}. So we have

aπ^{⟨ℓ⟩}m_{1}^{⟨ℓ⟩} = aπ^{⟨ℓ⟩}ι^{⟨ℓ⟩}m_{1}π^{⟨ℓ−1⟩}
= (a − ∑_{i∈[0,ℓ−1]} aπ^{⟨i⟩}ι^{⟨i⟩})m_{1}π^{⟨ℓ−1⟩}
= \underbrace{am_{1}π^{⟨ℓ−1⟩}}_{= 0 since am_{1} ∈ A^{⩽ℓ−2,z}} − ∑_{i∈[0,ℓ−1]} aπ^{⟨i⟩} \underbrace{ι^{⟨i⟩}m_{1}π^{⟨ℓ−1⟩}}_{= 0 since i ⩽ ℓ − 1}
= 0 .

Since A is diagonally resolving and since ℓ ⩾ 1, there exists a′ ∈ A^{⟨ℓ+1⟩,z[−2]} such that aπ^{⟨ℓ⟩} = a′m_{1}^{⟨ℓ+1⟩} = a′ι^{⟨ℓ+1⟩}m_{1}π^{⟨ℓ⟩} .

So

am_{1} = aπ^{⟨ℓ⟩}ι^{⟨ℓ⟩}m_{1} + (a − aπ^{⟨ℓ⟩}ι^{⟨ℓ⟩})m_{1}
= a′ι^{⟨ℓ+1⟩}m_{1}π^{⟨ℓ⟩}ι^{⟨ℓ⟩}m_{1} + (a − aπ^{⟨ℓ⟩}ι^{⟨ℓ⟩})m_{1}
= a′ι^{⟨ℓ+1⟩}\underbrace{m_{1}m_{1}}_{= 0} −( ∑_{i∈[0,ℓ−1]} \underbrace{a′ι^{⟨ℓ+1⟩}m_{1}π^{⟨i⟩}ι^{⟨i⟩}m_{1}}_{∈A^{⩽ℓ−1,z[−1]}m_{1}} ) + (\underbrace{a − aπ^{⟨ℓ⟩}ι^{⟨ℓ⟩}}_{∈A^{⩽ℓ−1,z[−1]}m_{1}})m_{1}
∈ A^{⩽ℓ−1,z[−1]}m_{1} .

This proves the claim. Suppose given z ∈ Mor(Z) and a ∈ A^{z[−1]} such that am_{1} ∈ A^{⩽k−1,z}. We have to show that am_{1} ∈^{!} A^{⩽k,z[−1]}m_{1} . There exists ℓ ∈ Z such that a ∈ A^{⩽ℓ,z[−1]}. If ℓ ⩽ k, we have am_{1} ∈ A^{⩽ℓ,z[−1]}m_{1} ⊆ A^{⩽k,z[−1]}m_{1} . If ℓ ⩾ k + 1, we obtain

am_{1} ∈ A^{⩽ℓ,z[−1]}m_{1} ∩ A^{⩽k−1,z}
= (A^{⩽ℓ}m_{1} ∩ A^{⩽k−1})^{z}
⊆^{Claim} (A^{⩽ℓ−1}m_{1} ∩ A^{⩽k−1})^{z}
⊆^{Claim} …
⊆^{Claim} (A^{⩽k}m_{1} ∩ A^{⩽k−1})^{z}
= (A^{⩽k}m_{1})^{z}
= A^{⩽k,z[−1]}m_{1} .

Proposition 60 Suppose given an A_{1}-algebra (A, (m_{1})) over Z. For z ∈ Mor(Z), suppose given an augmented projective resolution of the module (HA)^{z} (over R), written as follows.

⋯ → Ã^{⟨2⟩,z[−2]} ⟶^{d^{⟨2⟩,z[−2]}} Ã^{⟨1⟩,z[−1]} ⟶^{d^{⟨1⟩,z[−1]}} Ã^{⟨0⟩,z[0]} ⟶^{ε^{z}} (HA)^{z} → 0

These linear maps assemble to shift-graded linear maps between Z-graded modules as follows.

⋯ → Ã^{⟨2⟩} ⟶^{d^{⟨2⟩}} Ã^{⟨1⟩} ⟶^{d^{⟨1⟩}} Ã^{⟨0⟩} ⟶^{ε} HA → 0

Here d^{⟨i⟩} is of degree 1 for i ∈ Z_{⩾1} . Moreover, ε is of degree 0. Let Ã^{⟨i⟩} := 0 for i ∈ Z_{<0} . Let d^{⟨0⟩} := 0 : Ã^{⟨0⟩} → Ã^{⟨−1⟩} = 0, as shift-graded linear map of degree 1. Let Ã := ⨁_{i∈Z} Ã^{⟨i⟩}. So Ã = (Ã, (Ã^{⟨i⟩})_{i}) is a split-filtered Z-graded module. Then there exist • a shift-graded linear map e^{⟨k⟩} : Ã^{⟨k⟩} → Ã^{⩽k−2} of degree 1 for k ∈ Z_{⩾0} ,

• a shift-graded linear map q_{1} : Ã → A of degree 0 such that, letting the shift-graded linear map m̃_{1} : Ã → Ã of degree 1 be defined by

ι^{⟨i⟩} · m̃_{1} := d^{⟨i⟩} · ι^{⟨i−1⟩} + e^{⟨i⟩} · ι^{⩽i−2}

for i ∈ Z_{⩾0} , then the following assertions (1, 2, 3, 4, 5, 6) hold.

(1) Ã = (Ã, (m̃_{1}), (Ã^{⟨i⟩})_{i}) is a minimal eA_{1}-algebra over Z.

(2) (q_{1}) is a quasiisomorphism of A_{1}-algebras from Ã to A.

(3) Ã^{⟨i⟩,z} is a projective module (over R) for i ∈ Z and z ∈ Mor(Z).

(4) Ã is diagonally resolving.

(5) q_{1}|ZA ; Ã^{⟨0⟩} · ρ exists and is piecewise surjective.

(6) We have Ã^{⩽j}m̃_{1} = BÃ ∩ Ã^{⩽j−1} for j ∈ Z.

\xymatrix{ & Ã^{⟨i−1⟩,z[1]} &  & Ã^{⟨i−2⟩,z[1]} & … & Ã^{⟨0⟩,z[1]} \\  &  & … \ar@{-->}[urrr] \\ Ã^{⟨i⟩,z} \ar@{-->}[uur]^{d^{⟨i⟩,z}} \ar[uurrr]}
_{components of e^{⟨i⟩,z}}

In Kadeishvili’s Theorem 50, an A_{∞}-structure was constructed on the cohomology HA of a given A_{∞}-algebra A, in case R is a field. We will construct a minimal eA_{∞}-structure – so in particular, an A_{∞}-structure – on an arbitrarily chosen projective resolution Ã, in the sense of Proposition 60, of the cohomology HA of a given A_{∞}-algebra A; cf. Theorem refXXX below. Note that in case R is a field, one may choose the trivial projective resolution, where ε = id_{A} and where A^{⟨i⟩} = 0 for i ∈ Z_{⩾1} , so that Ã = HA. So in this case the assertions of said Theorems coincide; cf. Remark 57.

XXXAAAXXX

Appendix A

Problems and solutions

A.1 Problems

Problem 1 (Introduction) Consider the commutative ring Z. Consider the Z-algebra Z. Determine the isoclasses of the Z-modules M that have a chain of submodules

M = M_{0} ⊇ M_{1} ⊇ M_{2} ⊇ M_{3} = 0

such that

\begin{aligned}M_{0}/M_{1} & ≃ & Z/(2) \\ M_{1}/M_{2} & ≃ & Z/(4) \\ M_{2}/M_{3} & ≃ & Z/(2) . \end{aligned}

Problem 2 (§1.1.1) Let Cat denote the (1-)category of categories, (1-)morphisms being functors. Let Set denote the category of sets, morphisms being maps.

(1) Given a set X, how many isoclasses does the pair category X^{×2} have?

(2) Construct a full and faithful functor P : Set → Cat sending X to X^{×2}.

(3) Show that the functor Ob : Cat → Set has P as a right adjoint, i.e. Ob ⊣ P.

(4) Determine unit and counit of the adjunction in (3).

Problem 3 (§1.1.1) Let Poset denote the category of posets and monotone maps.

(1) Suppose given a poset X. Show that we have a subcategory CX of the pair category X^{×2} with Ob(CX) = X and Mor(CX) = { (x, y) ∈ X^{×2} : x ⩽ y }.

(2) Construct a functor C : Poset \overline{}Cat.

(3) Given n ∈ Z_{⩾0} , we write Δ_{n} := C[0, n].

We have the monotone map ω : [0, 1] → [0, n] , 0 ↦ 0, 1 ↦ n.

Suppose given a category Z and z ∈ Mor(Z). Let F_{z} : Δ_{1} → Z, (0, 1) ↦ z.

Let n ⩾ 1. Show that fact_{n}(z) is in bijection to { Δ_{n} ⟶^{G} Z : G is a functor such that G ∘ (Cω) = F_{z} } .

Problem 4 (§1.1.2) Let Z = (Z, S, deg) be a grading category. Show.

(1) The shift S is an automorphism of Z-grad_{0} .

(2) Z-grad is a category.

(3) By S(f, k) := (Sf, k) for (f, k) ∈ Mor(Z-grad), an automorphism S on Z-grad is defined.

(4) Z-grad_{0} is additive.

(5) Z-grad_{0} is isomorphic to a subcategory of Z-grad.

Is this subcategory full? Does Z-grad have a zero object?

Problem 5 (§1.1.1) Let Z = (Z, S, deg) and Z̃ = (Z̃, S̃, d̃eg) be grading categories. A (1-)morphism of grading categories from Z to Z̃ is a functor F : Z → Z̃ such that

\begin{aligned}F (zS) & = & (F z)S̃ \\ (z) deg & = & (F z)d̃eg \end{aligned}

for z ∈ Mor(Z).

(1) Show that grading categories, together with morphisms of such, form a category Grad.

(2) Show that (Z, S^{−}, − deg) is a grading category, where (S^{−})_{X,Y} := (S_{X,Y})^{−} for X, Y ∈ Ob(Z) and z(− deg) := −(z deg) for z ∈ Mor(Z). Construct an automorphism of order 2 on the category of grading categories.

(3) Show that (id_{X}) deg = 0 for X ∈ Ob(Z).

(4) Show that there exists exactly one morphism of grading categories from Z to Z, i.e. that Z is the terminal grading category.

(5) Show that there is a bijection from the set of morphisms of grading categories from Z to Z to the set of endomorphisms of Z of degree 0.

(6) Suppose given a morphism of grading categories Z ⟶^{F} Z̃.

Show that there exist functors

Z-grad [figure: F_{&}; F^{&}] Z̃-grad

having (F^{&}M̃)^{z} = M̃^{F z} for M̃ ∈ Ob(Z̃-grad) and z ∈ Mor(Z), having (F_{&}M)z̃ = ⨁_{\begin{matrix}z∈Mor(Z) \\ F z=z̃ \end{matrix}}M^{z} for M ∈ Ob(Z-grad) and z̃ ∈ Mor(Z̃) and having F_{&} ⊣ F^{&}.

Problem 6 (§1.1.3) Let (Z, S, deg) be a grading category.

Define a category (Z-grad)^{×n,±} such that we have a functor

(Z-grad)^{×n,±} \underrightarrow{⨂_{i∈[1,n]}} Z-grad
(L_{i} ⟶^{(f_{i},k_{i})} M_{i})_{i∈[1,n]} ⟼ (⨂_{i∈[1,n]} L_{i} \underrightarrow{⨂_{i∈[1,n]}(f_{i},k_{i})} ⨂_{i∈[1,n]} M_{i}) .

Problem 7 (§1.1.3) Let Z = (Z, S, deg) be a grading category.

Suppose given 1 ⩽ ℓ ⩽ n and Z-shift-graded linear maps L_{i} ⟶^{(f_{i},k_{i})} M_{i} for i ∈ [1, n].

Suppose given Z-shift-graded linear maps L ⟶^{(f,k)} M and L̃ ⟶^{(f̃,k̃)} M̃.

(1) Show that

(M_{1} ⊗ … ⊗ M_{ℓ}) ⊗ (M_{ℓ+1} ⊗ … ⊗ M_{n}) = M_{1} ⊗ … ⊗ M_{n} .

(2) Show that

((f_{1}, k_{1}) ⊗ … ⊗ (f_{ℓ}, k_{ℓ})) ⊗ ((f_{ℓ+1}, k_{ℓ+1}) ⊗ … ⊗ (f_{n}, k_{n})) = (f_{1}, k_{1}) ⊗ … ⊗ (f_{n}, k_{n}) .

(3) Construct a Z-graded module Ṙ such that (f, k) ⊗ (idṘ , 0) = (f, k) and (idṘ , 0) ⊗ (f, k) = (f, k).

(4) Construct an isomorphism L ⊗ L̃ ⟶_{∼}^{τ_{L,L̃}} L̃ ⊗ L in Z-grad, and likewise τ_{M,M̃} , such that the following quadrangle commutes.

\xymatrix{L ⊗ L̃ \ar[r]_{∼}^{τ_{L,L̃}} \ar[d]_{(f,k)⊗(f̃,k̃)} & L̃ ⊗ L \ar[d]^{(−1)^{kk̃}(f̃,k̃)⊗(f,k)} \\ M ⊗ M̃ \ar[r]_{∼}^{τ_{M,M̃}} & M̃ ⊗ M}

Problem 8 (Problem 17) Let B be an algebra.

(1) Let A be a linear additive category. Let N ⊆ A be a full additive subcategory. Write

Null_{A,N} (X, Y) := { X ⟶^{f} Y : there exists N ∈ Ob(N) and morphisms X ⟶^{u} N ⟶^{v} Y such that f = uv } .

Let A/N be the category that has

Ob(A/N) := Ob(A)
A/N(X, Y) := _{A}(X, Y)/ Null_{A,N} (X, Y) for X, Y ∈ Ob(A/N) .

For X ⟶^{f} Y ⟶^{g} Z in A, we define composition of the respective residue classes in A/N by

(f + Null_{A,N} (X, Y)) · (g + Null_{A,N} (Y, Z)) = f · g + Null_{A,N} (X, Z) .

Show that A/N is a linear additive category. Show that A ⟶^{R} A/N is a linear functor with RN ≃ 0 for N ∈ Ob(N).

We often write f̄:= f + Null_{A,N} (X, Y).

Given a linear additive category B and a linear functor A ⟶^{F} B with F N ≃ 0 for N ∈ Ob(N), show that there exists a unique linear functor A/N ⟶^{F̄} B such that F = F̄ ∘ R.

(2) Let A := C(B -Mod) be the category of complexes of B-modules. Let the differential of a complex X ∈ Ob(A) be denoted by d = d_{X} . Let N ⊆ A be the full additive subcategory of split acyclic complexes, i.e. those isomorphic to a complex of the

form ⋯ → U^{i−1} ⊕ U^{i} \underrightarrow{(\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix})} U^{i} ⊕ U^{i+1} → …, where U^{i} ∈ Ob A for i ∈ Z.

Show that Null_{A,N} (X, Y) consists of those morphisms of complexes X ⟶^{f} Y for

which there exists a tuple of morphisms (X^{i} ⟶^{h^{i}} Y^{i−1})_{i∈Z} such that

f^{i} = h^{i}d_{Y}^{i−1} + d_{X}^{i}h^{i+1} for i ∈ Z.

Define K(B -Mod) := A/N to be the homotopy category of complexes of B-modules. Write shorthand _{K}(X, Y) := _{K(B -Mod)}(X, Y) for X, Y ∈ Ob(K(B -Mod)) = Ob(C(B -Mod)).

(3) Let M be a B-module. Let P be a projective resolution of M with augmentation ε : P_{0} → M. Let Conc(M) ∈ Ob(C(B -Mod)) have M at position 0, and 0 elsewhere. Let ε̂: P → Conc(M) be the morphism of complexes having entry ε at position 0.

Let Q be a complex consisting of projective B-modules, bounded above. Show that K(Q, ¯ ε̂) : _{K}(Q, P) → _{K}(Q, Conc(M)) is an isomorphism.

(4) Using the universal property from (1), construct a shift functor S on K(B -Mod) such that (SX)^{i} = X^{i+1} and such that d_{SX}^{i} = −d_{X}^{i+1} for i ∈ Z. Show that S is an automorphism. We also write S^{k} =: (−)^{[k]} for k ∈ Z.

Problem 9 (Problem 17) Let q ∈ Z_{⩾1} . Consider the cyclic group C_{q} = ⟨ c : c^{q} ⟩. Abbreviate K := K(RC_{q} -Mod).

(1) Construct a projective resolution P of the trivial RC_{q}-module R that is periodic of period length 2.

(2) Calculate _{K}(P, Conc(R)^{[i]}) for i ∈ Z.

(3) Calculate _{K}(P, P^{[i]}) for i ∈ Z.

(4) Calculate the composition map

K(P, P^{[i]}) ⊗ _{K}(P^{[i]}, P^{[i+j]}) → _{K}(P, P^{[i+j]})

for i, j ∈ Z.

Problem 10 (§1.1.1) Let Z and Z̃ be grading categories. Let Z ^{\overline{}^{F}} Z̃ be a morphism of grading categories; cf. Problem 5. Let n ∈ Z_{⩾1} . Let M_{i} be a Z-graded module for i ∈ [1, n]. Write \underline{M} := (M_{i})_{i∈[1,n]} .

(1) Construct an isomorphism F_{&}(⨂_{i∈[1,n]} M_{i}) ⟶_{∼}^{σ_{\underline{M}}} ⨂_{i∈[1,n]} F_{&}M_{i} in Z̃-grad.

(2) Show that the following quadrangle commutes.

\xymatrix{F_{&}(⨂_{i∈[1,n]} M_{i}) \ar[r]^{σ_{\underline{M}}} \ar[d]_{F_{&}(⨂_{i∈[1,n]}(f_{i} , k_{i}))} & ⨂_{i∈[1,n]} F_{&}M_{i} \ar[d]^{⨂_{i∈[1,n]} F_{&}(f_{i} , k_{i})} \\ F_{&}(⨂_{i∈[1,n]} M′_{i}) \ar[r]^{σ_{\underline{M}′}} & ⨂_{i∈[1,n]} F_{&}M′_{i}}

Problem 11 (§1.2) Let Z be a grading category. Let A be a Z-graded module. Suppose given shift-graded maps m_{1} : A → A of degree 1 and m_{2} : A^{⊗2} → A of degree 0. For n ∈ Z_{⩾3} , we let m_{n} := 0, as shift-graded linear map from A^{⊗n} to A of degree 2 − n. Suppose that (m_{n})_{n∈Z_{⩾1}} satisfies the Stasheff equations for k ∈ [1, 3]. Suppose that for each X ∈ Ob(Z), there exists an element 1_{X} ∈ A^{id^{X}} such that for z, w ∈ Mor(Z) such that zt_{Z} = X = ws_{Z} and for a ∈ A^{z} and b ∈ A^{w}, we have (a ⊗ 1_{X})m_{2} = a and (1_{X} ⊗ b)m_{2} = b. Show that (A, (m_{n})_{n∈Z_{⩾1}}) is a differential graded algebra over Z.

Problem 12 (§1.2) Suppose given a grading category Z. Suppose given A_{∞}-algebras Ã and A. Suppose given a shift-graded linear map f_{1} : Ã → A of degree 0.

Suppose that f_{1}^{⊗k} · m_{k}^{A} = mÃ ; k · f_{1} for k ∈ Z_{⩾1} .

Let f_{k} = 0 for k ∈ Z_{⩾2} , as shift-graded linear map from Ã^{⊗k} to A of degree 1 − k. Show that (f_{k})_{k∈Z_{⩾1}} is a morphism of A_{∞}-algebras.

Problem 13 (§XXX) Let B be an algebra. Suppose given a diagram X′ ⟶^{i} X ⟶^{r} X′′ in C(B -Mod) such that X^{′k} ⟶^{i^{k}} X^{k} ⟶^{r^{k}} X^{′′k} is short exact for k ∈ Z. Such a diagram is called a short exact sequence of complexes in B.

(1) Suppose given T ⟶^{f} X in C(B -Mod) such that fr = 0. Show that there exists a unique morphism T ⟶^{f′} X′ such that f′i = f.

(2) Suppose given X ⟶^{g} T in C(B -Mod) such that ig = 0. Show that there exists a unique morphism X′′ ⟶^{g′′} T such that rg′′ = g.

(3) A Z-graded B-module M is a tuple M = (M^{z})_{z∈Z} of B-modules M^{z}. A graded B-linear map f : L → M between Z-graded B-modules is a tuple f = (f^{z})_{z∈Z} of B-linear maps f^{z}. Write B-Z-grad for the category of Z-graded B-modules and graded B-linear maps. Construct an additive functor H : C(B -Mod) → B-Z-grad having

(HX)^{k} = Kern(d^{k})/ Im(d^{k−1})

for a complex X with differential d = (X^{k} ⟶^{d^{k}} X^{k+1})_{k}.

For Y ⟶^{f} Z in C(B -Mod), we often write ((HY)^{k} ⟶^{(Hf)^{k}} (HZ)^{k}) =: (H^{k}Y ⟶^{H^{k}f} H^{k}Z).

(4) Construct a B-linear map H^{k}X′′ ⟶^{γ_{(i,r)}^{k}} H^{k+1}X′ for k ∈ Z, called connector of the given short exact sequence X′ ⟶^{i} X ⟶^{r} X′′, subject to the following conditions (i, ii).

(i) The sequence

… → H^{k}X′ ⟶^{H^{k}i} H^{k}X ⟶^{H^{k}r} H^{k}X′′ ⟶^{γ_{(i,r)}^{k}} H^{k+1}X′ ⟶^{H^{k+1}i} H^{k+1}X ⟶^{H^{k+1}r} H^{k+1}X′′ → …

is exact at each position.

(ii) Given a morphism of short exact sequences, i.e. a commutative diagram

\xymatrix{X′ \ar[r]^{i} \ar[d]^{f′} & X \ar[r]^{r} \ar[d]^{f} & X′′ \ar[d]^{f′′} \\ Y′ \ar[r]^{j} & Y \ar[r]^{s} & Y′′}

in C(B -Mod) with (i, r) and (j, s) short exact, we get, for k ∈ Z, the commutative quadrangle

\xymatrix{H^{k}X′′ \ar[r]^{γ_{(i,r)}^{k}} \ar[d]^{H^{k}f′′} & H^{k+1}X′ \ar[d]^{H^{k+1}f′} \\ H^{k}Y′′ \ar[r]^{γ_{(j,s)}^{k}} & H^{k+1}Y′ .}

Problem 14 (§1.3, §1.4) Suppose given an algebra B. Suppose given n ⩾ 1.

Suppose given X_{s} ∈ Ob C(B -Mod) for s ∈ [1, n]. Abbreviate \underline{X} := (X_{s})_{s∈[1,n]} .

Abbreviate Z := Z × [1, n]^{×2}, C := C(B -Mod) and K := K(B -Mod).

(1) Show that (Z Hom_{B}(\underline{X}))^{(j,(s,t))} = _{C}(X_{s}, X_{t}^{[j]}) for (j, (s, t)) ∈ Mor(Z).

(2) Show that (H Hom_{B}(\underline{X}))^{(j,(s,t))} = _{K}(X_{s}, X_{t}^{[j]}) for (j, (s, t)) ∈ Mor(Z).

(3) Show that m_{2}^{Hom_{B}(\underline{X})} induces a map m_{2}^{H Hom_{B}(\underline{X})} : H Hom_{B}(\underline{X})^{⊗2} → H Hom_{B}(\underline{X}) that maps [f] ⊗ [g] to [f · g] for each composable pair of morphisms (f, g) in C, where we use brackets to denote residue classes of morphisms of C in K.

Problem 15 (§1.4) Let Z be a grading category.

Let L and M be Z-graded modules.

Let L ⟶^{f} M be a shift-graded linear map of degree a ∈ Z.

(1) Let K := Kern(f), i.e. K^{z} := Kern(L^{z} ⟶^{f} M^{z[a]}) for z ∈ Mor(Z). Let K ⟶^{i} L be the shift-graded inclusion map of degree 0.

Suppose given a Z-graded module T and a shift-graded linear map T ⟶^{t} L of degree d such that tf = 0.

Show that there exist a unique shift-graded linear map T ⟶^{ť} K of degree d such that ťi = t.

(2) Suppose R to be a field. Suppose f to be piecewise surjective.

Show that there exists a piecewise injective shift-graded linear map L ⟵^{g} M of degree −a such that gf = id_{M} .

Problem 16 (§1.4) Let Z be a grading category. Let n ∈ Z_{⩾1} . Let K_{i}, L_{i} and M_{i} be Z-graded modules for i ∈ [1, n]. Let K_{i} ⟶^{u_{i}} L_{i} be a shift-graded linear map of degree c_{i} ∈ Z for i ∈ [1, n]. Let L_{i} ⟶^{f_{i}} M_{i} be a piecewise surjective shift-graded linear map of degree a_{i} ∈ Z for i ∈ [1, n].

(1) Show that ⨂_{i∈[1,n]} f_{i} is piecewise surjective.

(2) For i ∈ [1, n], suppose that K_{i} ⟶^{u_{i}} L_{i} ⟶^{f_{i}} M_{i} to be exact at L_{i}, i.e. suppose K_{i}^{z[−c_{i}]} ⟶^{u_{i}} L_{i}^{z} ⟶^{f_{i}} M_{i}^{z[a_{i}]} to be exact at L_{i}^{z} for z ∈ Mor(Z). Suppose given a Z-graded module T and a shift-graded linear map ⨂_{i∈[1,n]} L_{i} ⟶^{t} T of degree d. Suppose that (id^{⊗j−1} ⊗u_{j} ⊗ id^{⊗n−j})t = 0 for j ∈ [1, n]. Show that there exists a unique shift-graded linear map ⨂_{i∈[1,n]} M_{i} ⟶^{t̃} T of degree d − ∑_{i∈[1,n]} a_{i} such that (⨂_{i∈[1,n]} f_{i})t̃= t.

Problem 17 (§1.4) Let p > 0 be a prime. Let P ∈ Ob C(F_{p}C_{p} -Mod) be the projective resolution of the trivial F_{p}C_{p}-module as found in Problem 9.(1). Let \underline{X} := (P), so that \underline{X} has P as its only tuple entry. Let A := Hom_{F_{p}C_{p}}(\underline{X}) be the regular differential graded category, i.e. differential graded algebra over Z = Z × [1, 1]^{×2}. Recall from Problem 9 and Problem 14 that we have calculated the Z-graded module HA, i.e. that we know F_{p}-linear generators for its graded pieces. Find a minimal A_{3}-structure (m̃_{1}, m̃_{2}, m̃_{3}) on HA and a quasiisomorphism (q_{1}, q_{2}, q_{3}) : HA → A of A_{3}-algebras.

Problem 18 (§1.4) Suppose R to be a field. Let Z be a grading category. Let n ∈ [1, ∞]. Let A be a unital A_{n}-algebra over Z. Consider the shift-graded linear residue class map ZA ⟶^{ρ} HA of degree 0. Show that there exists a shift graded linear map ZA ⟵^{σ} HA of degree 0 such that σρ = id_{HA} and such that (1_{X}ρ)σ = 1_{X} for X ∈ Ob(Z).

Problem 19 (§1.6) Let Z be a grading category.

(1) Let Ṽ = (Ṽ , Δ̃) and V = (V, Δ) be coalgebras over Z. Let Ṽ ⟶^{f} V be a morphism of coalgebras. Suppose that f is piecewise bijective. Show that f is an isomorphism of coalgebras, i.e. that there exists a morphism of coalgebras Ṽ ⟵^{g} V such that fg = idṼ and gf = id_{V} . Then g is uniquely determined and written f^{−} := g.

(2) Let Ṽ = (Ṽ , Δ̃, δ̃) and V = (V, Δ, δ) be coalgebras with differential over Z. Let Ṽ ⟶^{f} V be a morphism of coalgebras with differential. Suppose that f is piecewise bijective. Show that f is an isomorphism of coalgebras with codifferential, i.e. that there exists a morphism of coalgebras with codifferential Ṽ ⟵^{g} V such that fg = idṼ and gf = id_{V} . Then g is uniquely determined and written f^{−} := g.

(3) Let Ṽ = (Ṽ , Δ̃) and V = (V, Δ) be coalgebras over Z.

Let Ṽ ⟶^{f} V be an isomorphism of coalgebras. Suppose given a codifferential δ on V. Show that fδf^{−} is a codifferential on Ṽ.

(4) Let V = (V, Δ) be a coalgebra over Z. Let λ : V \overline{}Ṙ be a shift-graded linear map of degree 0 ; cf. Problem 7.(3). Recall that Ṙ ⊗ V = V = V ⊗ Ṙ by identification. Let δ_{λ} := Δ(id ⊗λ) − Δ(λ ⊗ id). Show that δ_{λ} is a coderivation. Coderivations of this form are called inner.

Problem 20 (§1.1.2) Let Z be a grading category. Let I be a set. Let V_{i} be a Z-graded module for i ∈ I. Recall that the Z-graded module ⨁_{i∈I} V_{i} is defined by letting (⨁_{i∈I} V_{i})^{z} = ⨁_{i∈I} V_{i}^{z} for z ∈ Mor(Z).

(1) Given j ∈ I, construct a shift-graded linear inclusion map ι_{j} : V_{j} → ⨁_{i∈I} V_{i} of degree 0 and a shift-graded linear projection map π_{j} : ⨁_{i∈I} V_{i} → V_{j} of degree 0.

(2) Suppose given a Z-graded module S. Suppose given d ∈ Z. Suppose given a shift-graded linear map s_{j} : S → V_{j} of degree d for j ∈ I. Show that there exists a unique shift-graded linear map s : S → ⨁_{i∈I} V_{i} of degree d such that sπ_{j} = s_{j} for j ∈ I.

(3) Suppose I to be finite.

Suppose given a Z-graded module T. Suppose given d ∈ Z. Suppose given shiftgraded linear maps t_{j} : V_{j} → T of degree d for j ∈ I. Show that there exists a unique shift-graded linear map t : ⨁_{i∈I} V_{i} → T of degree d such that ι_{j}t = t_{j} for j ∈ I.

Problem 21 (§1.6) Let Z be a grading category. Let n ∈ [1, ∞] . Let (A, (m_{ℓ})_{ℓ}) be a pre-A_{n}-algebra.

Write 𝔪 := ((^{ω}m_{ℓ})_{ℓ∈[1,n]∩Z})β_{Coder,n,A^{[1]}} .

(1) Suppose given p ∈ [1, n]. Write 𝔪′ := ((^{ω}m_{ℓ})_{ℓ∈[1,p]∩Z})β_{Coder,p,A^{[1]}} .

Show that 𝔪′ = 𝔪|_{T_{⩽p}(A^{[1]})}^{T_{⩽p}(A^{[1]})} .

(2) Suppose n ∈ Z_{⩾1} . Suppose that (m_{ℓ})_{ℓ} satisfies the Stasheff equation at each k ∈ [1, n − 1].

Suppose given z ∈ Mor(Z) and a ∈ ((A^{[1]})^{⊗n})^{z}. Consider the following assertions.

(i) We have

a ( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = n \end{matrix}}(id^{⊗r} ⊗^{ω}m_{s} ⊗ id^{⊗t}) · ^{ω}m_{r+1+t}) = 0

(ii) We have a𝔪^{2} = 0.

Show that (i) and (ii) are equivalent.

Problem 22 (§1.6) Let Z be a grading category. Let n ∈ [1, ∞] . Let (Ã, (m̃_{ℓ})_{ℓ∈[1,n]∩Z}) and (A, (m_{ℓ})_{ℓ∈[1,n]∩Z}) be pre-A_{n}-algebras. Let f = (f_{ℓ})_{ℓ∈[1,n]∩Z} be a pre-A_{n}-morphism from Ã to A.

Write

𝔪̃ := ((^{ω}m̃_{ℓ})_{ℓ∈[1,n]∩Z})β_{Coder,n,Ã^{[1]}}
𝔪 := ((^{ω}m_{ℓ})_{ℓ∈[1,n]∩Z})β_{Coder,n,A^{[1]}}
𝔣 := ((^{ω}f_{ℓ})_{ℓ∈[1,n]∩Z})β_{Coalg,n,Ã^{[1]},A^{[1]}} .

(1) Suppose given p ∈ [1, n]. Write

𝔪̃′ := ((^{ω}m̃_{ℓ})_{ℓ∈[1,p]∩Z})β_{Coder,p,Ã^{[1]}}
𝔪′ := ((^{ω}m_{ℓ})_{ℓ∈[1,p]∩Z})β_{Coder,p,A^{[1]}}
𝔣′ := ((^{ω}f_{ℓ})_{ℓ∈[1,p]∩Z})β_{Coalg,p,Ã^{[1]},A^{[1]}}

Show that

\begin{aligned}𝔪̃′ & = & 𝔣|_{T_{⩽p}(Ã^{[1]})}^{T_{⩽p}(Ã^{[1]})} \\ 𝔪′ & = & 𝔪|_{T_{⩽p}(A^{[1]})}^{T_{⩽p}(A^{[1]})} \\ 𝔣′ & = & 𝔣|_{T_{⩽p}(Ã^{[1]})}^{T_{⩽p}(A^{[1]})} . \end{aligned}

(2) Suppose n ∈ Z_{⩾1} . Suppose that (f_{ℓ})_{ℓ} satisfies the Stasheff equation for morphisms at each k ∈ [1, n − 1].

Suppose given z ∈ Mor(Z) and ã ∈ ((Ã^{[1]})^{⊗n})^{z}. Consider the following assertions.

(i) We have

ã ( ∑_{\begin{matrix}(r,s,t)⩾(0,1,0) \\ r+s+t = n \end{matrix}}(id^{⊗r} ⊗^{ω}m̃_{s} ⊗ id^{⊗t}) · ^{ω}f_{r+1+t}) = ã ( ∑_{r∈[1,k]} ∑_{\begin{matrix}(i_{j})_{j∈[1,r]}⩾(1)_{j} \\ ∑_{j} i_{j} = n \end{matrix}}( ⨂_{j∈[1,r]} ^{ω}f_{i_{j}}) · ^{ω}m_{r})

(ii) We have ã(𝔪̃𝔣 − 𝔣𝔪) = 0.

Show that (i) and (ii) are equivalent.

Problem 23 (§XXX) Let n ∈ [1, ∞]. Let Z be a grading category.

Let (A, (m_{ℓ})_{ℓ}), (A′, (m′_{ℓ})_{ℓ}), (A′′, (m′′_{ℓ})_{ℓ}), (A′′′, (m′′′_{ℓ})_{ℓ}) be A_{n}-algebras over Z.

Let f = (f_{ℓ})_{ℓ} : A → A′, f′ = (f′_{ℓ})_{ℓ} : A′ → A′′, f′′ = (f′′_{ℓ})_{ℓ} : A′′ → A′′′ be A_{n}-morphisms.

Write ^{ω}f := (^{ω}f_{ℓ})_{ℓ}.

Define

f · f′ := ^{ω^{−}}(((^{ω}f)β · (^{ω}f′)β)α) .

(1) Show that f · f′ is a morphism of A_{n}-algebras from A to A′′.

(2) Write f · f′ in terms of (f_{ℓ})_{ℓ} and (f′_{ℓ})_{ℓ}. What is the entry of f · f′ at ℓ = 1 ?

(3) Show that (f · f′) · f′′ = f · (f′ · f′′).

(4) Suppose given shift-graded linear maps g : A → A′ and g′ : A′ → A′′ of degree 0. Define strict_{n}(g) := (g, 0, 0, …). When is strict_{n}(g) : A → A′ a morphism of A_{n}-algebras? Is strict_{n}(id_{A}) a morphism of A_{n}-algebras? If strict_{n}(g) and strict_{n}(g′) are morphisms of A_{n}-algebras, show that strict_{n}(gg′) = strict_{n}(g) · strict_{n}(g′).

(5) Show that f · strict_{n}(id_{A′}) = f and that strict_{n}(id_{A′}) · f′ = f′.

(6) Define the category A_{n}-Z-alg of A_{n}-algebras over Z and A_{n}-morphisms. Therein, define the subcategory strict-A_{n}-Z-alg of A_{n}-algebras over Z and strict A_{n}-morphisms.

(7) Show that H is a functor from A_{n}-Z-alg to Z-grad_{0} .

Problem 24 (§XXX) Let n ∈ [1, ∞]. Let Z ⟶^{F} Z̃ be a morphism of grading categories; cf. Problem 5. Let (A, (m_{k})_{k}) be an A_{n}-algebra over Z.

(1) Show that F_{&}A = (F_{&}A, (σ^{−} ·F_{&}m_{ℓ})_{ℓ}) is an A_{n}-algebra over Z̃, where σ = σ_{(A,...,A)} ; cf. Problem 10.

(2) Consider the case n = ∞, u ∈ Z_{⩾1} , Z = Z × [1, u]^{×2}, Z̃ = Z and P being the projection, mapping a morphism (j, (s, t)) to j.

Given a unital Z × [1, u]^{×2}-algebra A, i.e. an A_{∞}-category with set of objects [1, u], show that its total A_{∞}-algebra P_{&}A is unital.

Problem 25 (§XXX) Let Z be a grading category.

Let A = (A, (m_{1}), (A^{⟨i⟩})_{i}) be a minimal eA_{1}-algebra over Z.

Suppose that there exist shift-graded linear map d^{⟨i⟩} : A^{⟨i⟩} → A^{⟨i−1⟩} of degree 1 and shift-graded linear map e^{⟨i⟩} : A^{⟨i⟩} → A^{⩽i−2} of degree 1 for i ∈ Z_{⩾0} such that

ι^{⟨i⟩} · m_{1} = d^{⟨i⟩} · ι^{⟨i−1⟩} + e^{⟨i⟩} · ι^{⩽i−2} .

holds for i ∈ Z_{⩾0} .

(1) Express the Stasheff equation at 1 in terms of d^{⟨i⟩} and e^{⟨i⟩}, where i ∈ Z_{⩾0} .

(2) Show that A is diagonally resolving if and only if Kern(d^{⟨i⟩}) = Im(d^{⟨i+1⟩}) for i ∈ Z_{⩾1} .

Problem 26 (§XXX) Let Z be a grading category.

Suppose given an eA_{∞}-algebra (A, (m_{k})_{k} , (A^{⟨i⟩})_{i}) over Z. Suppose that A^{⟨i⟩} = 0 for i ∈ Z ∖ [0, ℓ].

For which integers k ∈ Z_{⩾1} is the Schmid condition on m_{k} not void?

For which integers k ∈ Z_{⩾1} is the strong Schmid condition on m_{k} not void?

(1) Consider the case ℓ = 1.

(2) Consider the case ℓ = 2.

(3) Consider the case ℓ = 3.

Problem 27 (§XXX) Let Z be a grading category.

Suppose given an eA_{∞}-algebra (A, (m_{k})_{k}, (A^{⟨i⟩})_{i}) over Z. Let k ⩾ 1. Let (j_{1}, …, j_{k}) ∈ Z_{⩾0}^{×k}.

What bound results from the Schmid condition for the image of A^{⟨j_{1}⟩} ⊗ … ⊗ A^{⟨j_{k}⟩} under a summand of the Stasheff equation at k?

Problem 28 Let X = (X, ⩽) be a poset. We call X artinian if it does not contain a strictly descending chain. We call X superartinian if X_{⩽ξ} is finite for all ξ. We call X discrete if (⩽) = (=). We call X narrow if each discrete subposet of X is finite.

Suppose given k ∈ Z_{⩾1} and posets Y_{1} , …, Y_{k}.

(1) Show that X is artinian if and only if each nonempty subposet of X has a minimal element.

(2) If X is superartinian, show that X is artinian. Does the converse hold?

(3) Construct the product ∏_{i∈[1,k]} Y_{i} in Poset, which is to be equipped with monotone

maps ∏_{i∈[1,k]} Y_{i} ⟶^{π_{j}} Y_{j} for j ∈ [1, k] such that for each poset T and each tuple

(T ⟶^{t_{i}} Y_{i})_{i} of monotone maps, there exists a unique monotone map T ⟶^{t} ∏_{i∈[1,k]} Y_{i} such that t · π_{j} = t_{j} for j ∈ [1, k].

(4) If Y_{i} is artinian for i ∈ [1, k], show that ∏_{i∈[1,k]} Y_{i} is artinian.

(5) If Y_{i} is superartinian for i ∈ [1, k], show that ∏_{i∈[1,k]} Y_{i} is superartinian.

(6) Show that Z_{⩾0}^{×k} := ∏_{i∈[1,k]} Z_{⩾0} is superartinian and narrow.

A.2 Solutions

Literatur

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[3] Sagave, S., DG-algebras and derived A-infinity algebras, Journal für die reine und angewandte Mathematik 639, p. 73–105, 2010.

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