Lecture 5: Künneth formula, Simplicial objects, Dold-Kan correspondence

Teacher: Benjamin Hennion

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Recall from last time:

$A$ commutative ring $M ⊆ Mod_A$

\[φ_1, …, φ_n= M ⟶ M \text{ st } φ_i φ_j = φ_j φ_i\] \[K(M, φ_\bullet): \\ ⋯ ⟶ \bigwedge^p A^n ⊗_A M ⟶ ⋯ ⟶ \bigwedge^1 A^n ⊗_A M ⟶ M\]

Def: $(φ_1, …, φ_n)$ regular sequence iff

\[∀i, \quad φ_i: M/(\Im φ_1 + ⋯ + \Im φ_{i-1}) ⟶ M/(\Im φ_1 + ⋯ + \Im φ_{i-1})\]

is injective

Prop: If $(φ_1, …, φ_n)$ is a regular sequence, then \(H^i(K(M, φ_\bullet)) = 0 \quad \text{ if } i≠0\\ H^0(K(M, φ_\bullet)) = M/(\Im φ_1 + ⋯ + \Im φ_n)\\\)

Lemma: Let $φ_{≤i} ≝ (φ_1, …, φ_i)$.

$φ_{i+1}$ induces a morphism of complexes

\[K(M, φ_{≤i}) \overset{\overline φ_{i+1}}{⟶} K(M, φ_{≤i})\]

We have an isomorphism

\[K(M, φ_{≤i}) ≅ Mc(\overline φ_{i+1})\]

Reminder: $Mc(f)^p = X^{p+1} ⊕ φ^p$ if $f: X ⟶ φ$

\[\begin{xy} \xymatrix{ Mc(\overline φ_{i+1})^{-(p+1)}: & \bigwedge^p A^i ⊗_A M \ar[rd]_{ϕ_p ∧ e_{i+1} ⊗ id_M} & ⊕ & \bigwedge^{p+1} A^i ⊗_A M \ar[ld]^{ϕ_{p+1} ∧ φ_{i+1}} \\ K(M, φ_{≤i+1}): & & \bigwedge^{p+1} A^{i+1} ⊗_A M } \end{xy}\] \[ϕ: A^i \hookrightarrow A^{i+1} \ni e_{i+1}\\ ϕ_p: \bigwedge^p A^i ⟶ \bigwedge^p A^{i+1}\]

+ check the differential

We get an exact sequence:

\[0 ⟶ K(M, φ_{≤ i}) ⟶ K(M, φ_{≤ i+1}) ⟶ K(M, φ_{≤ i})[1] ⟶ 0\]

Taking cohomology, we find a long exact sequence:

\[H^p(K(M, φ_{≤ i})) \overset{\overline φ_{i+1}}{⟶} H^p(K(M, φ_{≤ i})) ⟶ H^p(K(M, φ_{≤ i+1})) ⟶ H^{p+1}(K(M, φ_{≤ i}))\]

Assume that we know

\[H^p(K(M, φ_{≤ i})) = 0 \text{ for } p ≠0\]

then we find

\[H^p(K(M, φ_{≤ i+1})) = 0 \text{ for } p ≠ 0, -1\] \[\underbrace{H^{-1}(K(M, φ_{≤ i}))}_{=0} ⟶ H^{-1}(K(M, φ_{≤ i+1})) ⟶ \underbrace{H^0(K(M, φ_{≤ i})) \overset{\overline φ_{i+1}}{⟶} H^0(K(M, φ_{≤ i}))}_{≃ M/(\Im φ_1 + ⋯ + \Im φ_{i}) \overset{φ_{i+1} = f_{i+1}}{⟶} M/(\Im φ_1 + ⋯ + \Im φ_{i}) ≃}\]

Since the sequence $(φ_1, …, φ_n)$ is regular, the map $f_{i+1}$ is injective, and its kernel $H^{-1}(K(M, φ_{≤i+1})) = 0$.

By induction (check the case $n=0$), we are done!

Examples: $M = A = k[x_1, …, x_n]$.

\[φ_i: \begin{cases} M=A &⟶ M=A \\ 1 &⟼ x_i \end{cases}\]

NB: this map is no longer injective in the example where $ε^2=0$

Claim: $(φ_1, …, φ_n)$ is a regular sequence.

$K(A, φ_\bullet)$ is a projective resolution of $k$ as a $k[x_1, …, x_n]$-module.

In particular,

\[Tor_p^A(k, k) = 0 \text{ if } p ≥ n+1\]

NB: In general, finding a resolution of $A/I$ as a $A$-module is not so easy, and not necessarily finite ($k[ε] ⟶ k$)

Dual version: $\bigwedge^p A^n ≃ \bigwedge^{n-p} A^n$

\[\underbrace{M ⟶ \bigwedge^1 A^n ⊗ M}_{m \; ⟼ \; \sum (-1)^{?} e_i ⊗ φ_i (m)} ⟶ \bigwedge^2 A^n ⊗ M\]

⟹ can write the de Rham complex.

VIII. Künneth formula

$𝒞 ≝ Mod_A$, $A$ commutative ring

comes with a tensor product.

Def: Let $X^\bullet, Y^\bullet ∈ C(𝒞)$ be complexes over $𝒞$.

We define

\[(X^\bullet ⊗ Y^\bullet)^n ≝ \bigoplus_{p+q=n} X^p ⊗ Y^q\]

We define a differential:

\[\begin{cases} (X^\bullet ⊗ Y^\bullet)^n &⟶ (X^\bullet ⊗ Y^\bullet)^{n+1} \\ \underbrace{x}_{∈ X^p} ⊗ \underbrace{y}_{∈ Y^q} &⟼ \underbrace{d_x x ⊗ y}_{∈ X^{p+1} ⊗ Y^q} + \underbrace{(-1)^p x ⊗ d_y y}_{∈ X^p ⊗ Y^{q+1}} \end{cases}\]

Exercise: Check that $d^2=0$ and why $(-1)^p$ is needed

Künneth formula

Theorem (Künneth formula): Assume $A = k$ is a field, $φ^\bullet$ is bounded ($φ^n = 0$ if $n » 0$ ou $n « 0$).


\[H^n(X^\bullet ⊗_k Y^\bullet) ≃ \bigoplus_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^\bullet)\]


  • $M, N$ two manifolds \(H_{dR}^n(M × N) ≃ \bigoplus_{p+q=n} H^p_{dR}(M) ⊗_ℝ H^q_{dR}(N)\)

  • Find the example that is closer to what you study…

Def: Let $X^\bullet, Y^\bullet ∈ C(𝒞)$.

The cross product is the family of maps:

\[\begin{cases} H^p(X^\bullet) ⊗_A H^q(Y^\bullet) &⟶ H^{p+q}(X^\bullet ⊗_A Y^\bullet) \\ \underbrace{[α]}_{∈ H^p(X^\bullet)} ⊗ \underbrace{[β]}_{∈ H^q(Y^\bullet)} &⟼ [α ⊗ β] \end{cases}\]

where $α ∈ Z^p(X^\bullet), \; β ∈ Z^q(Y^\bullet)$.

(we have to check this is well defined)

Proof (of the formula):

  1. $k$ is a field ⟹ for all $V ∈ Vect_k$, $- ⊗_k V$ is exact.

    So for all $V ∈ Vect_k$, \(H^p(X^\bullet ⊗_A V) ≃ H^p(X^\bullet) ⊗_k V\)

  2. $Y^\bullet$ is bounded, we can assume (modulo some shifting):

    \[Y^\bullet = 0 ⟶ Y^0 ⟶ Y_1 ⟶ ⋯ ⟶ Y^n ⟶ 0\]


    \[\begin{xy} \xymatrix{ Y^{≤i} & 0 \ar[r] & Y^0 \ar[r] & ⋯ \ar[r] & Y^i \ar[r] & 0 \\ Y^{≤i+1} & 0 \ar[r] & Y^0 \ar[r] \ar[u] & ⋯ \ar[r] & Y^i \ar[r] \ar[u] & Y^{i+1} \ar[u] \ar[r] & 0 \\ & & & & 0 \ar[r] & Y^{i+1} \ar[r]\ar[u] & 0 \\ } \end{xy}\]

    We find a short exact sequence:

    \[0 ⟶ Y^{i+1}[-(i+1)] ⟶ Y^{≤ i+1} ⟶ Y^{≤i} ⟶ 0 \quad ⊛\]
  3. Using 1. and 2., prove the Künneth formula by induction on the length of $Y^\bullet$:

    Use the cross product to find a morphism

    \[\bigoplus_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^\bullet) ⟶ H^n(X^\bullet ⊗_k Y^\bullet)\]

    The functor

    \[Y ⟼ \bigoplus_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^\bullet)\]

    is exact:

    \[\underbrace{0}_{\text{ if } q ≠ i+1} ⟶ H^q(Y^{≤ i+1}) \underbrace{≃}_{\text{ if } q ≠ i, i+1} H^q (Y^{≤ i}) ⟶ \underbrace{0}_{\text{ if } q ≠ i}\] \[0 ⟶ H^i(Y^{≤ i+1}) ⟶ H^i(Y^{≤i}) ⟶ Y^{i+1} ⟶ H^{i+1}(Y^{≤i+1}) ⟶ H^{i+1}(Y^{≤i}) = 0\]

    Assume that I know that

    \[\bigoplus_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^{≤i}) ⟶ H^n(X^\bullet ⊗_k Y^{≤ i})\]

    is an isomorphism.


    \[\begin{xy} \xymatrix{ \bigoplus\limits_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^{i+1}[-(i+1)]) \ar[r]^-{\sim \text{ by 1.}} \ar[d] & H^n(X^\bullet ⊗_k Y^{i+1}[-(i+1)]) \ar[d] \\ \bigoplus\limits_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^{≤i+1}) \ar[r]^-{f} \ar[d] & H^n(X^\bullet ⊗_k Y^{≤ i+1}) \ar[d] \\ \bigoplus\limits_{p+q=n} H^p(X^\bullet) ⊗_k H^q(Y^{≤i}) \ar[r]^-{ \sim } & H^n(X^\bullet ⊗_k Y^{≤ i})\\ \underbrace{H^{n-i}(X^\bullet) ⊗ Y^{i+1}}_{\text{ if } q = i} \ar[r]^-{ \sim } & \underbrace{\qquad}_{\text{build an exact sequence}} } \end{xy}\]

    $q=i+1$ or $q=i$ ⟹ you can complete the exact sequence.

    Use the five lemma to show that $f$ is an iso.

NB: the boundedness assumption can be removed. Moreover, you only need the $Y^i$’s to be flat and the differential preserves flatness.

2. Algebras

Assume $X^\bullet ∈ C(𝒞)$ with a (bilinear) product

\[X^\bullet ⊗ X^\bullet \overset{μ}{⟶} X^\bullet\]

Then, using the cross product, we find:

\[H^p(X^\bullet) ⊗ H^q(X^\bullet) ⟶ H^{p+q}(X^\bullet ⊗ X^\bullet) ⟶ H^{p+q}(X^\bullet)\\ [α] ⊗ [β] ⟼ [α] ∪ [β]\]

This is called the cup product.

If $μ$ is an associative, then so is the cup product.

NB: Just because you have a cup product doesn’t mean that you have such a $μ$ map.

Example: $M$ manifold.

$Ω_M^\bullet$ de Rham complex has a multiplication.

\[ω_1, ω_2 ⟼ ω_1 ∧ ω_2\]

⟹ cup product on $H_{dR}^\bullet(M)$

Question: What does it mean for $μ$ to be commutative?

\[V ∈ Vect_k, \quad μ: V ⊗ V ⟶ V\]

$μ$ commutative ⟺

\[\begin{xy} \xymatrix{ V ⊗ V \ar[r]^{μ} \ar[d]_{τ}^\sim & V \\ V ⊗ V \ar@{.>}[ru]_{ μ } & } \end{xy}\]

commutes, where $τ(v ⊗ w) = w ⊗ v$

Question: $X^\bullet ∈ C(𝒞)$.

\[τ: X^\bullet ⊗ X^\bullet \overset{\sim}{⟶} X^\bullet ⊗ X^\bullet\] \[\bigoplus_{p+q=n} X^p ⊗ X^q = (X^\bullet ⊗ X^\bullet)^n \overset{τ}{⟶} (X^\bullet ⊗ X^\bullet)^n = \bigoplus_{p+q=n} X^p ⊗ X^q\]

The naive $τ$ does not preserve the differential!

Better one:

\[τ_{pq}: \begin{cases} X^p ⊗ X^q &⟶ X^p ⊗ X^q \\ x ⊗ y &⟼ (-1)^{pq} y ⊗ x \end{cases}\] \[τ ≝ \bigoplus τ_{pq}: X^\bullet ⊗ X^\bullet \overset{\sim}{⟶} X^\bullet ⊗ X^\bullet\]

Consequence: A commutative algebra in complexes over $A$ is a complex $X^\bullet$, with maps

\[X^p ⊗ X^q \overset{μ_{pq}}{⟶} X^{p+q}\]

that are “associative” and such that

\[μ_{pq}(x ⊗ y) = (-1)^p μ_{qp}(y ⊗ x)\]

elements of odd degrees anti-commute.

Ex: $Ω_M^\bullet$, and in general any $\bigwedge^\bullet E$.

Can we derive

\[\begin{cases} CAlg_k &⟶ Vect_k \\ A &⟼ \Ker(A ⊗ A ⟶ A) \end{cases}\]

? We don’t have an additive category (can we add algebra morphisms? No! the unit will be mapped to $2$)

Same: a functor on the category of manifolds, topological spaces, etc… can’t be derived with what we’ve done. Same for the homotopy groups $π_i$’s!

But actually, there’s a way to make sense of “deriving” this functor.

Simplicial methods

1. Simplicial objects

Idea: They look like triangulations

\[\begin{xy} \xymatrix{ Δ_1, Δ_2 \ar@<+3pt>[r] \ar@<-3pt>[r] \ar[r] & \text{ 3 edges } \ar@<+3pt>[r] \ar@<-3pt>[r] & \text{ 1 vertex} } \end{xy}\]


(cf picture)

$X$ algebraic variety/manifold

\[\bigcup_i U_i ⟶ X\]

an open covering (finite, and there are finitely many connected components in intersections).

Denote by

\[U_{i_1, …, i_p} = U_{i_1} ∩ ⋯ ∩ U_{i_p}\] \[\begin{xy} \xymatrix{ \bigcup U_{i_1, i_2, i_3} \ar@<+3pt>[r] \ar@<-3pt>[r] \ar[r] & \bigcup_{i,j} U_{i,j} \ar@<+3pt>[r] \ar@<-3pt>[r] & \bigcup_{i} U_i \ar[r] & X } \end{xy}\]

⟹ Cech cohomology

Def: Let $Δ$ be the category whose objects are $\lbrace [n] ≝ \lbrace 0 < 1 < ⋯ < n\rbrace, n ∈ ℕ \rbrace$ and morphisms are $[n] ⟶ [m]$ are order preserving maps.

A simplicial object in any category $𝒞$ is a functor

\[A: Δ^{op} ⟶ 𝒞\]

We write $A_n$ for $A([n])$.

We denote by $s𝒞$ the category of simplicial objects in $𝒞$ (morphisms are natural transformations).


\[[n+1]: \quad 0 < 1 < ⋯ < i-1 < i < i+1 < ⋯ < n+1\\ [n]: \quad 0 < 1 < ⋯ < i-1 < i < ⋯ < n\]

Let $ε_i$ be the map:

\[ε_i: \begin{cases} [n] &⟶ [n+1] \\ j ≤ i-1 &⟼ j \\ j > i-1 &⟼ j+1 \\ \end{cases}\]


\[η_i: \begin{cases} [n+1] &⟶ [n] \\ j ≤ i &⟼ j \\ j > i &⟼ j-1 \\ \end{cases}\]

The $ε_i$’s are called faces, the $η_i$’s are called degeneracies.

There are relations:

\[ε_j ε_i = ε_i ε_{j-1} \quad \text{ if } i <j \qquad [n] ⟶ [n+1]\\ η_j η_i = η_i η_{j+1} \quad \text{ if } i ≤ j \qquad [n+2] ⟶ [n]\\ η_j ε_i = \begin{cases} ε_i η_{j-1} &&\text{ if } i<j \\ id &&\text{ if } i=j \text{ or } i=j+1\\ ε_{i-1} η_j &&\text{ if } i > j+1 \end{cases}\]

NB: Any map $[n] ⟶ [m]$ in $Δ$ factorizes uniquely into

\[[n] \overset{\text{surj composed of } η_i \text{'s}}{⟶} [p] \overset{\text{incl composed of } ε_i \text{'s}}{⟶} [m]\]

Prop: A simplicial object $A_\bullet$ is determined by

  • the objects $A_n$
  • the maps $A(ε_i) ≝ \partial_i: A_{n+1} ⟶ A_n$

    (those are called the face maps)

  • the maps $A(η_i) ≝ σ_i: A_n ⟶ A_{n+1}$

    (those are called the degeneracy maps)

satisfying the relations:

\[\partial_i \partial_j = \partial_{j-1} \partial_i \quad \text{ if } i < j\\ σ_i σ_j = σ_{j+1} σ_i \quad \text{ if } i < j\\ \partial_i σ_j = \begin{cases} σ_{j-1} \partial_i &&\text{ if } i < j \\ id &&\text{ if } i = j \text{ or } i= j+1 \\ σ_j \partial_{i-1} &&\text{ if } i > j+1 \\ \end{cases}\]

Example: for categories (nerve), degeneracies correspond to identity morphisms.

(cf picture)


Nerve of $f$


\[U = \bigcup_i U_i \overset{f}{⟶} X\]

We can define the nerve of $f$:

\[N(f)_p ≝ \underbrace{U ×_X U ×_X ⋯ ×_X U}_{\text{fiber product } p \text{ times}}\]

It is a simplicial object in $Var$/$Mnfld$/$𝒞$.

Singular simplices in a topological space $X$


\[\lbrace x_0, …, x_n \; \mid \; 0 ≤ x_i < 1, \sum\limits_{ i } x_i = 1 \rbrace = Δ^n ⊆ ℝ^n\]

(standard simplices)

Any morphism $[n]⟶[p]$ induces $Δ^n ⟶ Δ^p$. We get a functor

\[Δ^\bullet: Δ ⟶ Top\]

$X$ a topological space.

\[S_\bullet (X): \begin{cases} Δ^{op} &⟶ Set \\ [n] &⟼ \Hom_{Top}(Δ^n, X) \end{cases}\]

“singular simplices in $X$”

2. Dold-Kan correspondence

Assume that $𝒞$ is abelian.

Until now, we’ve been studying cochain complexes:

\[X^\bullet \qquad d: X^p ⟶ X^{p+1}\]

There are also chain complexes:

\[Y_\bullet \qquad d: Y_p ⟶ Y_{p-1}\]

You can go from one to the other by setting $Y_p ≝ X^{-p}$

Denote by $Ch(𝒞){≥0}$ the category of chain complexes such that $Y_p = 0 \text{ if } p < 0$. If $Ch(𝒞){≥0} \ni Y_\bullet$, we denote by $H_n(Y_\bullet)$ the homology of $Y_\bullet$


Let $A_\bullet ∈ s𝒞$.

Moore complex

Define the Moore complex of $A$ as the chain complex $C(A_\bullet)$ given by

\[C(A_\bullet)_n ≝ A_n\] \[d ≝ \sum\limits_{ i=0 }^n (-1)^i \partial_i \quad : C(A_\bullet)_n = A_n ⟶ C(A_\bullet)_{n-1} = A_{n-1}\]

Check that $d^2 = 0$.

Normalized complex

Define the normalized complex of $A_\bullet$:

\[\begin{xy} \xymatrix{ N(A_\bullet) = \bigcap\limits_{i=0}^{n-1} \ker (\partial_i) \ar@{^{(}->}[r] \ar[d]_{ d = (-1)^n \partial_n} & C(A_\bullet)_n \ar[d]^{ d } \\ N(A_\bullet)_{n-1} \ar@{^{(}->}[r] & C(A_\bullet)_{n-1} } \end{xy}\]

$N(A_\bullet)$ is a subcomplex of $C(A_\bullet)_\bullet$.


$X$ topological space, $S_\bullet(X): Δ^{op} ⟶ Set$.

$A$ commutative ring.

\[F_A: \begin{cases} Set &⟶ Mod_A \\ I &⟼ \underbrace{A^{(I)}}_{\text{free } A \text{-mod with basis } I} \end{cases}\] \[F_A \circ S_\bullet(X): Δ^{op} ⟶ Mod_A ∈ sMod_A \overset{C(-)_\bullet}{⟶} Ch(A)_{≥0}\]

$H_\bullet(C(F_A \circ S_\bullet(X)))$ is called the singular homology of $X$ with coefficients in $A$. $H^{sing}(X,A)$.

  • Let $A$ be a $k \text{-}CAlg$, $x_1, …, x_n ∈ A, \; x_i: A ⟶ A$

    $(\bigwedge^p A^n)$ semi-simplicial (i.e. no $σ_i$’s) $A$-module

    \[C_\bullet(\bigwedge^\bullet A^n) ≃ K(A, x_\bullet)\]
  • Bar resolution ($ℤ$ as a $ℤ[G]$-module)

NB: There is a dual version of all that, taking to co-simplicial objects to cochain complexes:

\[\underbrace{\bigwedge^p A^n}_{\text{cosimplicial}} ⟶ \underbrace{C(𝒞)^{≥0}}_{\text{cochain complexes}}\]


  • Cech cohomology: Way to compute the cohomology of a sheaf: $X$ algebraic variety over $k$, $U_i \underbrace{\hookrightarrow}_{\text{open}} X$. Let $F ∈ Sh(X, k)$ such that

    • $U_i$ is affine
    • $F$ is quasi coherent
    \[H^n(U_i, F) = 0 \text{ if } n ≠ 0\]

    Assume also that

    \[∀i,j, \quad U_{ij} = U_i ∩ U_j \text{ is affine}\] \[Δ^{op} ⟶ Var\\ \begin{xy} \xymatrix{ \bigsqcup_{i,j,k} U_{ijk} \ar@<+3pt>[r] \ar@<-3pt>[r] \ar[r] & \bigsqcup_{i,j} U_{ij} \ar@<+3pt>[r] \ar@<-3pt>[r] & \bigcup_{i} U_i } \end{xy}\\ \begin{xy} \xymatrix{ F(𝒰) = \prod\limits_{ i } F(U_i) \ar@<+3pt>[r] \ar@<-3pt>[r] & \bigsqcup_{i,j} F(U_{ij}) \ar@<+3pt>[r] \ar@<-3pt>[r] \ar[r] & \bigsqcup_{i,j,k} F(U_{ijk}) } \end{xy}\\\]

    cosimplicial object in $Vect_k$.

    $C^\bullet(F(𝒰)) ∈ C(𝒞)$ is called the Cech complex.

    Cech cohomology:
    \[H^n(C^\bullet(F(𝒰))) ≃ H^n(X, F)\]

Prop: The monomorphism

\[N(A_\bullet)_\bullet \hookrightarrow C(A_\bullet)_\bullet\]

is a quasi-isomorphism ($A_\bullet$ is a simplicial object)

Proof: Assume $𝒞 = Mod_A$

Let $D(A_\bullet)_n$ be generated by the images of the $σ_i$’s ($σ_i: A_{n-1} ⟶ A_n$)

\[D(A_\bullet)_n ⊆ C(A_\bullet)_n = A_n\]


  • $D(A_\bullet)$ is a subcomplex of $C(A_\bullet)$

  • \[C(A_\bullet)_n ≃ N(A)_n ⊕ D(A_\bullet)_n\]
  • $D(A_\bullet)$ is acyclic

Dold-Kan correspondence

$𝒞$ abelian category. The functor

\[N_\bullet: s𝒞 ⟶ Ch(𝒞)_{≥0}\]

is an equivalence of categories.

Its inverse $K$ is given by:

\[K(X_\bullet)_n = \bigoplus\limits_{[n] ↠ [p]} X_p\]

(cf picture for the definition of $σ_i$ and $\partial_i$)

NB: We can do in $s𝒞$ what we can do in $Ch(𝒞)_{≥ 0}$ (or in $C(𝒞)$).

  • \[π_i \leftrightarrow H_i \leftrightarrow H^{-i}\]
  • homotopies $\leftrightarrow$ homologies
  • $⊗ \leftrightarrow ⊗$

NB: Bar construction: (trivial) group action on the point.

Sketch of the proof:

$X_\bullet ∈ Ch(𝒞)_{≥ 0}$.

  1. \[X_\bullet \overset{α}{≅} N_\bullet(K_\bullet(X_\bullet))\]

    functorial in $X$.

    By definition:

    \[N_n = \bigcap\limits_{i=0}^{n-1} \ker \partial_i\]
  2. \[\Hom_{s𝒞}(K(X_\bullet), A_\bullet) \overset{\text{bijection}}{⟶} \Hom_{Ch(𝒞)_{≥0}}(NK(X_\bullet), N(A_\bullet)) ≅ \Hom_{Ch(𝒞)_{≥0}}(X_\bullet, N(A_\bullet))\]


    \[K_\bullet ⊣ N_\bullet\]

    Consider the counit:

    \[K_\bullet N_\bullet \overset{θ}{⇒} id\]

    So the composition

    \[N_\bullet \overset{α_{N_\bullet(-)}}{⇒} N_\bullet K_\bullet N_\bullet \overset{N_\bullet (θ)}{⇒} N_\bullet\]

    is the identity, as we have an adjunction.

    In particular, for all $A ∈ s𝒞$:

    \[N_\bullet K_\bullet N_\bullet (A) ≅ N_\bullet(A)\]

    Def: We say that $N_\bullet$ is conservative if it detects isomorphisms, i.e. $f: A → B$ is an iso iff $N_\bullet f$ is an iso.

    Claim: $N_\bullet$ is conservative


    \[KN(A_\bullet) ≅ A_\bullet\]

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