Cohomology

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Cohomology, as its name suggests, can be regarded as the dual concept to homology. Yet if the \(k\)th cohomology \(H^k(X)\) of a space \(X\) were simply the dual of the \(k\)th homology \(H_k(X)\), there would be little reason to study it separately.

In fact, cohomology furnishes a more refined invariant than homology: for instance, it carries a natural product structure, and two spaces with the same homology need not be homotopic if their product structures differ. In this post we review the definition of cohomology and its basic properties.

The Universal Coefficient Theorem for Homology

Before turning to the main discussion, we first recall homology with coefficients, which we treated after §Computation of Homology, ⁋Definition 6 (Eilenberg-Steenrod axioms). In defining simplicial or singular homology, we observed that by replacing the chain groups \(C_\bullet(X)\) or \(C_\bullet^\Delta(X)\) with their tensor product over \(\mathbb{Z}\) with an abelian group \(A\), we obtain chain complexes

\[C_\bullet(X;A):=C_\bullet(X)\otimes_\mathbb{Z}A,\qquad C_\bullet^\Delta(X;A):=C_\bullet^\Delta(X)\otimes_\mathbb{Z}A\]

and from these we define homology groups

\[H_k(X;A),\qquad H_k^\Delta(X;A)\]

To compute these in practice, one might hope that \(H_k(X;A)\) or \(H_k^\Delta(X;A)\) could be expressed simply as \(H_k(X)\otimes_\mathbb{Z}A\) or \(H_k^\Delta(X)\otimes_\mathbb{Z}A\); however, since the tensor product is right-exact but not left-exact in general, this would be too much to expect. Instead, we must additionally measure the information lost by tensoring.

To this end, consider a chain complex \(C_\bullet\) of free abelian groups and the short exact sequence in \(\Ch_{\geq 0}(\Ab)\)

\[0 \rightarrow Z_\bullet \rightarrow C_\bullet \rightarrow B_{\bullet-1}\rightarrow 0\tag{1}\]

Here \(Z_k=\ker(\partial:C_k \rightarrow C_{k-1})\) and \(B_{k-1}=\im(\partial:C_k \rightarrow C_{k-1})\), the first map is the inclusion, and the second is the boundary map \(\partial\).

Since \(Z_{k}\) and \(B_{k-1}\) are subgroups of the free abelian groups \(C_k\) and \(C_{k-1}\), they are themselves free; in particular the third term \(B_{k-1}\) is free, so assigning a preimage to each basis element defines a section \(B_{k-1}\rightarrow C_k\), and the short exact sequence (1) is split exact. Therefore, for any abelian group \(A\), the sequence

\[0 \rightarrow Z_\bullet\otimes_\mathbb{Z}A \rightarrow C_\bullet\otimes_\mathbb{Z}A \rightarrow B_{\bullet-1}\otimes_\mathbb{Z}A\rightarrow 0\]

is also a split short exact sequence: the splitting amounts to a direct sum decomposition \(C_\bullet\cong Z_\bullet\oplus B_{\bullet-1}\), and tensoring commutes with direct sums, so the decomposition survives \(\otimes_\mathbb{Z}A\). Expanding these out gives a commutative diagram of the form

and thus by [Homological Algebra] §Long Exact Sequences, ⁋Theorem 1 we obtain the long exact sequence

\[\cdots B_k\otimes_\mathbb{Z}A\overset{\delta_k}{\longrightarrow}Z_k\otimes_\mathbb{Z}A\rightarrow H_k(C\otimes A)\rightarrow B_{k-1}\otimes_\mathbb{Z}A\overset{\delta_{k-1}}{\longrightarrow} Z_{k-1}\otimes_\mathbb{Z}A\rightarrow\cdots\tag{2}\]

To extract information from this exact sequence we must examine the connecting maps \(\delta\). Tracing through the definition, \(\delta_k:B_k\otimes_\mathbb{Z}A\rightarrow Z_k\otimes_\mathbb{Z}A\) is exactly \(-\otimes \id_A\) applied to the inclusion homomorphism \(i_k:B_k \rightarrow Z_k\). Thus, consider the short exact sequence

\[0 \rightarrow B_k\overset{i_k}{\longrightarrow} Z_k \overset{p_k}{\longrightarrow}H_k(C)\rightarrow 0\]

Since \(B_k\), \(Z_k\), and \(0\) are all free abelian groups, this may be viewed as a free resolution of \(H_k(C_\bullet)\), and therefore by definition we can embed \(\delta_k\) into the exact sequence

\[0 \rightarrow \Tor_1^\mathbb{Z}(H_k(C))\rightarrow B_k\otimes_\mathbb{Z}A\overset{\delta_k}{\longrightarrow} Z_k\otimes_\mathbb{Z}A\rightarrow H_k\otimes_\mathbb{Z}A\rightarrow 0\]

That is, we have isomorphisms

\[\ker\delta_{k-1}\cong \Tor_1^\mathbb{Z}(H_{k-1}(C), A),\qquad \coker\delta_k\cong H_k(C)\otimes_\mathbb{Z} A\]

and substituting these into (2) yields the short exact sequence

\[0 \rightarrow H_k(C)\otimes_\mathbb{Z}A\rightarrow H_k(C;A)\rightarrow \Tor_1^\mathbb{Z}(H_{k-1}(C), A)\rightarrow 0\]

On the other hand, because (1) is a split short exact sequence, we may choose a retraction \(r_k:C_k \rightarrow Z_k\). ([Multilinear Algebra] §Exact Sequences, ⁋Proposition 10) With this choice, \((p_k\circ r_k)\otimes \id_A\) induces a map \(H_k(C;A)\rightarrow H_k(C)\otimes_\mathbb{Z} A\) that serves as a retraction of the map \(H_k(C)\otimes_\mathbb{Z}A\rightarrow H_k(C;A)\) above. Hence we obtain the following.

Proposition 1 (Universal coefficient theorem for homology) For any topological space \(X\) and abelian group \(A\), there exists a short exact sequence

\[0 \rightarrow H_k(X)\otimes_\mathbb{Z}A\rightarrow H_k(X;A)\rightarrow \Tor_1^\mathbb{Z}(H_{k-1}(X), A)\rightarrow 0\]

Moreover, this sequence splits (non-canonically), and thus yields a (non-canonical) isomorphism

\[H_k(X;A)\cong \left(H_k(X)\otimes_\mathbb{Z}A\right)\oplus \Tor_1^\mathbb{Z}(H_{k-1}(X), A)\]

Definition of Cohomology and the Universal Coefficient Theorem

Just as in §Computation of Homology, ⁋Definition 6 (Eilenberg-Steenrod axioms), we define the Eilenberg-Steenrod axioms for cohomology; a contravariant functor together with connecting morphisms satisfying these axioms will be called a cohomology theory. Writing this out explicitly gives the following.

Definition 2 (Eilenberg-Steenrod axioms) For contravariant functors \(H^k\) from the category of pairs of topological spaces to the category of abelian groups, together with natural transformations

\[\delta: H^k(X) \rightarrow H^{k+1}(X,A)\]

the Eilenberg-Steenrod axioms are the following.

• (Homotopy) If two maps \((X,A) \rightarrow (Y,B)\) are homotopic, then the induced homomorphisms \(H^k(Y,B) \rightarrow H^k(X,A)\) coincide.
• (Excision) For \((X,A,Z)\) satisfying the conditions of §Computation of Homology, ⁋Theorem 2 (Excision theorem), the inclusion \((X\setminus Z, A\setminus Z)\hookrightarrow (X,A)\) induces an isomorphism.
• (Dimension) For the one-point space \(\ast\), \(H^k(\ast)=0\) for all \(k>0\).
• (Additivity) If \(X=\coprod X_\alpha\), then \(H^k(X)\cong\bigoplus H^k(X_\alpha)\).

• (Exactness) For each pair \((X,A)\) and the two inclusions \((A,\emptyset) \hookrightarrow (X,\emptyset)\) and \((X,\emptyset)\hookrightarrow (X,A)\), we have the long exact sequence

  \[\cdots \rightarrow H^k(X,A)\rightarrow H^k(X) \rightarrow H^k(A) \rightarrow H^{k+1}(X,A)\rightarrow \cdots\]

To establish the existence of a cohomology theory satisfying these axioms, we proceed as in §Homology and consider the chain complex of singular simplices of a topological space \(X\)

\[C_\bullet(X):\qquad\cdots \rightarrow C_{k+1}(X)\rightarrow C_k(X) \rightarrow C_{k-1}(X)\rightarrow \cdots\]

Fixing an abelian group \(A\) as the coefficient group, we may consider the following cochain complex, which corresponds to the dual of this chain complex:

\[(C^\vee)^\bullet(X;A):\qquad\cdots \leftarrow \Hom_\mathbb{Z}(C_{k+1}(X), A)\leftarrow\Hom_\mathbb{Z}(C_k(X),A)\leftarrow\Hom_\mathbb{Z}(C_{k-1}(X),A)\leftarrow\cdots\]

By [Algebraic Structures] §Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Theorem 6, this may be identified with

\[\qquad \cdots\leftarrow\Hom_A(C_{k+1}(X;A),A)\leftarrow \Hom_A(C_k(X;A),A)\leftarrow \Hom_A(C_{k-1}(X;A),A)\leftarrow\cdots\]

and thus regarded as the dual of the chain complex \(C_\bullet(X;A)\). We define the \(k\)th homology of this cochain complex \((C^\vee)^\bullet(X;A)\) as

\[H^k(X;A):=H_k(C^\vee)\]

and call it the \(k\)th cohomology of \(X\). The reason for using superscripts on \(H\) and \(C^\vee\) to index the grading is that, in contrast to homology, the long exact sequence runs in the direction of increasing index; henceforth, whenever there is no risk of confusion, we shall write \((C^\vee)^\bullet(X;A)\) as \(C^\bullet(X;A)\).

We must now examine the relationship between \(H^k(X;A)\) and \(H_k(X)\) thus defined. As noted at the outset, it is not the case that \(H^k(X;A)\cong H_k(X)^\ast\) holds in general. Nevertheless, by an argument parallel to the proof of Proposition 1 (Universal coefficient theorem for homology) above, we obtain the following.

Proposition 3 (Universal coefficient theorem for cohomology) For any topological space \(X\) and abelian group \(A\), there exists a short exact sequence

\[0\rightarrow\Ext_\mathbb{Z}^1(H_{k-1}(X), A)\rightarrow H^k(X;A)\rightarrow \Hom_\mathbb{Z}(H_k(X),A)\rightarrow 0\]

Moreover, this sequence splits (non-canonically), and thus yields a (non-canonical) isomorphism

\[H^k(X;A)\cong \Hom_\mathbb{Z}(H_k(X),A)\oplus \Ext^1_\mathbb{Z}(H_{k-1}(X),A)\]

Roughly speaking, this can be viewed as translating Proposition 1 (Universal coefficient theorem for homology) via [Algebraic Structures] §Abelian Groups, ⁋Theorem 15.

de Rham Cohomology

Associating a chain complex \(C_\bullet(X)\) to a topological space \(X\) amounts to encoding information about the subspaces of \(X\) in algebraic terms. In defining cohomology we apply \(\Hom_\mathbb{Z}(-,A)\) to \(C_\bullet(X)\) and take the homology of the resulting cochain complex; here

\[C^k(X;A)=\Hom_\mathbb{Z}(C_k(X), A)\]

and an arbitrary element may be thought of as a function assigning an element of \(A\) to each element of \(C_k(X)\) (that is, to each \(k\)-chain). Thus cohomology is essentially the study of functions defined on a space.

More concretely, for any \(c\in C_k(X)\) and \(\varphi\in C^k(X;A)\), we have the canonical pairing

\[C_k(X)\times C^k(X;A)\rightarrow A;\qquad (c,\varphi)\mapsto \varphi(c)\in A\]

If we write \(\partial\) for the boundary map of \(C_\bullet(X)\) and \(\delta\) for the coboundary map of \(C^\bullet(X;A)\) induced from it, then for any \(c\in C_{k+1}(X)\) and \(\varphi\in C^k(X;A)\) the relation

\[\langle \partial c, \varphi\rangle=\langle c, \delta\varphi\rangle\]

holds, and from this we obtain a pairing at the homology and cohomology level

\[H_k(X)\times H^k(X;A)\rightarrow A\]

¹

For example, to consider de Rham cohomology, let

\[\Omega^k(\mathbb{R}^n)=\{\text{$k$-forms on $\mathbb{R}^n$}\}\]

be the \(\mathbb{R}\)-vector spaces of differential \(k\)-forms. Here the coboundary map \(\Omega^k(\mathbb{R}^n)\rightarrow \Omega^{k+1}(\mathbb{R}^n)\) is given by the exterior derivative, and a differential \(k\)-form produces a number via integration when evaluated on a \(k\)-dimensional subset. Moreover, closed \(k\)-forms are the kernel of this coboundary, and exact \(k\)-forms are its image.

For instance, consider the differential \(2\)-form on \(\mathbb{R}^3\)

\[\omega=dx\wedge dy\]

A \(2\)-dimensional subset of \(\mathbb{R}^3\) is given by a map from a (unit) rectangle in \(\mathbb{R}^2\) into \(\mathbb{R}^3\), and through this we understand what it means to evaluate \(\omega\) on such a subset.

Suppose, for example, that the set

\[S = \{ (x, y, 0) \mid 0 \leq x \leq 1,\, 0 \leq y \leq 1 \}\]

is given. Then the integral of \(\omega\) over this set is computed simply as

\[\int_S \omega = \int_{x=0}^{1} \int_{y=0}^{1} 1\,dy\,dx = 1\]

As another example, if the surface

\[\Sigma = \{ (x, y, z) \mid x^2 + y^2 + z^2 = 1,\ z \geq 0 \}\]

is given, we first parametrize it as a map from \([0,\pi/2]\times[0,2\pi]\) to \(\Sigma\) using spherical coordinates

\[x = \sin \phi \cos \theta,\qquad y = \sin \phi \sin \theta,\qquad z = \cos \phi\]

and then, using \(dx \wedge dy = \sin \phi \cos \phi\, d\phi \wedge d\theta\), we compute the integral as

\[\begin{align*} \int_{\Sigma} \omega &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \sin \phi \cos \phi\, d\phi \, d\theta = \int_{0}^{2\pi} d\theta \int_{0}^{\pi/2} \sin \phi \cos \phi\, d\phi \\ &= 2\pi \times \frac{1}{2} \int_{0}^{\pi/2} \sin(2\phi) d\phi = 2\pi \times \frac{1}{2} \left[ -\frac{1}{2} \cos(2\phi) \right]_{0}^{\pi/2} \\ &= 2\pi \times \frac{1}{2} \left( -\frac{1}{2} [\cos(\pi) - \cos(0)] \right) = 2\pi \times \frac{1}{2} \left( -\frac{1}{2}(-1 - 1) \right) = 2\pi \times \frac{1}{2} \times 1 \\ &= \pi \end{align*}\]

Thus the differential \(2\)-form \(\omega\) can be regarded as a function that takes \(2\)-dimensional subsets such as \(S\) or \(\Sigma\) and outputs a number.

Now by the Poincaré lemma, any closed \(k\)-form on \(\mathbb{R}^n\) is always the exterior derivative of some suitable \((k-1)\)-form. Therefore, for any \(k>0\)

\[H^k_\dR(\mathbb{R}^n)=0\]

and for \(k=0\), the functions whose derivative vanishes are exactly the constant functions, so

\[H^0_\dR(\mathbb{R}^n)=\mathbb{R}\]

The de Rham cohomology defined in this way also satisfies all the conditions of Definition 2 (Eilenberg-Steenrod axioms), and hence by the uniqueness of cohomology theories (together with the fact that any singular chain can be approximated by a smooth chain), singular cohomology with \(\mathbb{R}\) coefficients and de Rham cohomology coincide. The computation above is then nothing more than translating the computation of §Homology, ⁋Proposition 11 into \(\mathbb{R}\)-valued cohomology via Proposition 3 (Universal coefficient theorem for cohomology).

Coefficients of (Co)homology

The de Rham cohomology examined above is an example of a cohomology theory whose coefficient group is not \(\mathbb{Z}\). Unlike singular or simplicial cohomology, de Rham cohomology naturally has \(\mathbb{R}\) as its coefficient group by definition.

Such a cohomology theory enjoys favorable properties: for instance, since \(\mathbb{R}\) is a torsion-free abelian group, \(\Tor_1^\mathbb{Z}(A,\mathbb{R})=0\) for any abelian group \(A\), and thus by Proposition 1 (Universal coefficient theorem for homology) we have the isomorphism

\[H_k(X;\mathbb{R})\cong H_k(X)\otimes_\mathbb{Z}\mathbb{R}\]

Moreover, since \(\mathbb{R}\) is an injective \(\mathbb{Z}\)-module, \(\Ext_\mathbb{Z}^1(A,\mathbb{R})=0\) for any abelian group \(A\), and hence this time Proposition 3 (Universal coefficient theorem for cohomology) yields the isomorphism

\[H^k(X;\mathbb{R})\cong \Hom_\mathbb{Z}(H_k(X),\mathbb{R})\]

Investigating this type of homology and cohomology then becomes a further topic of interest. To this end, recalling the chain complexes used to define \(H_k(X;A)\) and \(H^k(X;A)\), we observe that the two chain complexes

\[C_\bullet(X;A):=C_\bullet(X)\otimes_\mathbb{Z}A,\qquad C_\bullet^\Delta(X;A):=C_\bullet^\Delta(X)\otimes_\mathbb{Z}A\]

are, when \(A\) is a ring, chain complexes of \(A\)-modules, and the \(C^\bullet(X;A)\) defined above is likewise. Therefore, taking homology or cohomology of these yields \(A\)-modules as well.

On the other hand, we know that if \(A\) is a principal ideal domain, then any submodule of a free \(A\)-module is again free. Looking back at the proof of Proposition 1 (Universal coefficient theorem for homology), we see that it relied on the fact that \(\mathbb{Z}\) is a principal ideal domain, so every submodule of a free \(\mathbb{Z}\)-module (that is, of a free abelian group) is again free. Building on this, we can generalize the preceding two propositions as follows.

Theorem 4 (Universal coefficient theorem for homology, general version) Let \(A\) be a principal ideal domain, \(C_\bullet\) a chain complex of free \(A\)-modules, and \(M\) an arbitrary \(A\)-module. Then there exists a short exact sequence

\[0 \rightarrow H_k(C)\otimes_AM\rightarrow H_k(C\otimes_AM)\rightarrow \Tor_1^A(H_{k-1}(C), A)\rightarrow 0\]

Moreover, this sequence splits (non-canonically), and thus yields a (non-canonical) isomorphism

\[H_k(C\otimes_AM)\cong \left(H_k(C)\otimes_AM\right)\oplus \Tor_1^A(H_{k-1}(C), M)\]

Theorem 5 (Universal coefficient theorem for cohomology, general version) Let \(A\) be a principal ideal domain, \(C_\bullet\) a chain complex of free \(A\)-modules, and \(M\) an arbitrary \(A\)-module. Then there exists a short exact sequence

\[0\rightarrow\Ext_A^1(H_{k-1}(C), M)\rightarrow H_k(\Hom_A(C,M))\rightarrow \Hom_A(H_k(C),M)\rightarrow 0\]

Moreover, this sequence splits (non-canonically), and thus yields a (non-canonical) isomorphism

\[H_k(\Hom_A(C,M))\cong \Hom_A(H_k(C),M)\oplus \Ext^1_A(H_{k-1}(C),M)\]

The Mayer-Vietoris Sequence

Among the axioms of Definition 2 (Eilenberg-Steenrod axioms), the excision axiom allows us to compute the cohomology of a large space from that of smaller ones. The following proposition is the cohomology version of [Algebraic Topology] §Computation of Homology, ⁋Proposition 7; its proof consists of repeating the passage from [Algebraic Topology] §Computation of Homology, ⁋Definition 6 to [Algebraic Topology] §Computation of Homology, ⁋Proposition 7, now starting from Definition 2 (Eilenberg-Steenrod axioms).

Proposition 6 (Mayer-Vietoris sequence) Let a topological space \(X\) be expressed as the union \(X=U\cup V\) of two open sets, and consider a cohomology theory \(H\) defined on it. Then there exists a long exact sequence

\[\cdots \to H^{n}(X) \xrightarrow{(i^\ast, j^\ast)} H^{n}(U) \oplus H^{n}(V) \xrightarrow{k^\ast - l^\ast} H^{n}(U \cap V) \xrightarrow{\delta} H^{n+1}(X) \to \cdots\]

where \(i^\ast, j^\ast, k^\ast, l^\ast\) are the maps induced by the inclusions

\[i:U\rightarrow X,\quad j:V\rightarrow X,\quad k:U\cap V\rightarrow U,\quad l:U\cap V \rightarrow V\]

Tensor Product of Chain Complexes

The Mayer-Vietoris sequence lets us compute the homology or cohomology of a large space from that of its small subspaces. On the other hand, we can also form a larger space \(X\times Y\) as the product of two spaces \(X\) and \(Y\); the Künneth formula helps us compute the homology and cohomology of such a product space. To this end, we must first define the tensor product of two chain complexes \(C_\bullet\) and \(D_\bullet\).

Definition 7 Let \(A\) be a ring and \(C_\bullet, D_\bullet\) chain complexes of \(A\)-modules. Their tensor product \((C\otimes D)_\bullet\) is defined in degree \(k\) by

\[(C\otimes D)_k=\bigoplus_{p+q=k}C_p\otimes_A D_q\]

and the differential is defined on homogeneous elements by

\[\partial(x,y)=(\partial^Cx,y)+(-1)^{\deg(x)}(x,\partial^Dy)\]

and then extended linearly.

That is, \((C\otimes D)_\bullet\) can be regarded as the total complex of a double complex whose \((p,q)\)-component is \(C_p\otimes D_q\), with horizontal differential \(\partial^C\otimes\id_D\) and vertical differential \(\id_C\otimes \partial^D\). (§Homology, ⁋Definition 5)

The algebraic content of the Künneth formula is then captured by the following lemma.

Lemma 8 Let \(A\) be a principal ideal domain and \(C_\bullet\), \(D_\bullet\) chain complexes of \(A\)-modules, and suppose \(C_\bullet\) is a chain complex of free \(A\)-modules. Then for any \(k\) there exists a short exact sequence

\[0 \rightarrow \bigoplus_{p+q=k}H_p(C)\otimes_AH_q(D)\rightarrow H_k(C\otimes D)\rightarrow \bigoplus_{p+q=k-1}\Tor_1^A(H_p(C),H_q(D))\rightarrow 0\]

Moreover, this short exact sequence splits (non-canonically), and thus there is an isomorphism

\[H_k(C\otimes D)\cong \left( \bigoplus_{p+q=k}H_p(C)\otimes_AH_q(D)\right)\oplus \left(\bigoplus_{p+q=k-1}\Tor_1^A(H_p(C),H_q(D)) \right)\]

Proof

First, consider the short exact sequence

\[0 \rightarrow Z_p(C) \rightarrow C_p\rightarrow B_{p-1}(C)\rightarrow 0\]

Since \(B_{p-1}(C)\) and \(Z_p(C)\) are submodules of the free \(A\)-modules \(C_{p-1},C_p\) and \(A\) is a principal ideal domain, they are again free \(A\)-modules. Therefore, tensoring this short exact sequence with \(D_q\) yields the short exact sequence

\[0\rightarrow Z_p(C)\otimes D_q \rightarrow C_p\otimes D_q \rightarrow B_{p-1}(C)\otimes D_q\rightarrow 0\]

From the definition of the chain complex \((C\otimes D)_\bullet\), considering such short exact sequences for all \((p,q)\) with \(p+q=k\) and then taking the direct sum, we obtain the short exact sequence

\[0 \rightarrow (Z(C)\otimes D)_k \rightarrow (C\otimes D)_k \rightarrow (B(C)\otimes D)_{k-1}\rightarrow 0\]

Now considering the long exact sequence in homology arising from this short exact sequence, we obtain

\[\cdots \rightarrow H_{k}(B(C)\otimes D)\overset{\delta_k}{\longrightarrow} H_{k}(Z(C)\otimes D)\rightarrow H_{k}(C\otimes D)\rightarrow H_{k-1}(B(C)\otimes D)\overset{\delta_{k-1}}{\longrightarrow} H_{k-1}(Z(C)\otimes D)\rightarrow \cdots\]

In particular, focusing on \(H_k(C\otimes D)\), we obtain the short exact sequence

\[0 \rightarrow \coker\delta_k\rightarrow H_k(C\otimes D)\rightarrow \ker\delta_{k-1}\rightarrow 0 \tag{$\ast$}\]

To analyze \(\coker\delta_k\) and \(\ker\delta_{k-1}\), consider the short exact sequence

\[0 \rightarrow B_\bullet(C)\rightarrow Z_\bullet(C)\rightarrow H_\bullet(C)\rightarrow 0\]

and the exact sequence obtained by taking the tensor product with \(H_\bullet(D)\):

\[0 \rightarrow \Tor_1^A(H(C), H(D))_\bullet\rightarrow (B(C)\otimes H(D))_\bullet\rightarrow (Z(C) \otimes H(D))_\bullet \rightarrow (H(C)\otimes H(D))_\bullet \rightarrow 0\]

Here the initial \(0\) comes from \(Z_\bullet(C)\) being free. On the other hand, since free modules are flat, tensoring with a free module commutes with taking homology; hence in the above sequence we have

\[(B(C)\otimes H(D))_\bullet\cong H_\bullet(B(C)\otimes D)\qquad (Z(C)\otimes H(D))_\bullet \cong H_\bullet(Z(C)\otimes D)\]

The map \((B(C)\otimes H(D))_\bullet \rightarrow (Z(C)\otimes H(D))_\bullet\) arises from the inclusion \(B_\bullet(C)\rightarrow Z_\bullet(C)\), and after the above identification it coincides with \(\delta_\bullet\). Therefore we obtain

\[\coker \delta_k\cong (H(C)\otimes H(D))_k,\qquad \ker \delta_{k-1}\cong \Tor_1^A(H(C),H(D))_{k-1}\]

As for the splitting claim, since

\[0 \rightarrow Z_\bullet(C)\rightarrow C_\bullet \rightarrow B_{\bullet-1}(C) \rightarrow 0\]

is a split exact sequence, a section \(B_{\bullet-1}(C)\rightarrow C_\bullet\) induces a splitting of (\(\ast\)).

The Eilenberg-Zilber Theorem and the Künneth Formula

With the result of Lemma 8 in mind, our task is clear. Given two topological spaces \(X,Y\) and the corresponding chain complexes \(C_\bullet(X),C_\bullet(Y)\), we examine the relationship between the homology \(H_\bullet(X\times Y)\) of the product space \(X\times Y\) and the tensor product \((H(X)\otimes H(Y))_\bullet\) of the two chain complexes \(H_\bullet(X)\), \(H_\bullet(Y)\). The following theorem shows that these two algebraic objects are one and the same.

Theorem 9 (Eilenberg-Zilber) For two topological spaces \(X,Y\) and the chain complexes \(C_\bullet(X),C_\bullet(Y)\), and \(C_\bullet(X\times Y)\) obtained from them, there exists a chain homotopy equivalence between \((C(X)\otimes C(Y))_\bullet\) and \(C_\bullet(X\times Y)\); consequently

\[H_\bullet(C(X\times Y))\cong H_\bullet(C(X)\otimes C(Y))\]

holds.

This is usually proved using the acyclic models theorem, but the acyclic models theorem is in fact closer to a generalization of the Eilenberg-Zilber theorem, so invoking it here feels somewhat excessive. However, since a direct proof of the Eilenberg-Zilber theorem is quite tedious, we shall content ourselves with examining the two maps that appear in its proof:

\[\AW:C_\bullet(X\times Y) \rightarrow (C(X)\otimes C(Y))_\bullet,\qquad \EZ:(C(X)\otimes C(Y))_\bullet \rightarrow C_\bullet(X\times Y)\]

The proof of the acyclic models theorem is relegated to a separate post so as not to interrupt the flow.

First, the Alexander-Whitney map \(\AW:C_\bullet(X\times Y) \rightarrow (C(X)\otimes C(Y))_\bullet\) sends an arbitrary \(k\)-simplex \(\sigma:\Delta^k \rightarrow X\times Y\) to

\[\sum_p (\pi_X\circ \sigma\vert_{[v_0,\ldots,v_p]})\otimes (\pi_Y\circ \sigma\vert_{[v_p,\ldots v_k]})\in \bigoplus_{p+q=k}C_p(X)\otimes C_q(Y)\]

If \(X=Y\), this becomes the map that makes \(C(X)\) into a (differential graded counital coassociative) coalgebra via \(C(X)\rightarrow C(X)\otimes C(X)\), and for this reason it will reappear in the next post.

The Eilenberg-Zilber map \(\EZ:(C(X)\otimes C(Y))_\bullet \rightarrow C_\bullet(X\times Y)\) is defined on simple tensors by

\[\EZ(\sigma\otimes\tau)=\sum_{\substack{\alpha_1<\cdots <\alpha_p\\\beta_1<\cdots <\beta_q}}\sgn(\alpha_1,\ldots,\alpha_p,\beta_1,\ldots,\beta_q)(\sigma\circ s_{\alpha_p}\cdots s_{\alpha_1})\times(\tau\circ s_{\beta_q}\cdots s_{\beta_1})\]

which looks complicated as a formula, but is nothing more than the \(h_n\) function appearing in the proof of §Homotopy, ⁋Proposition 6—that is, the method of decomposing the prism \(\Delta^p\times \Delta^q\) into simplices. The conclusion of Theorem 9 (Eilenberg-Zilber) then follows from the two identities

\[\AW\circ\EZ=\id_{(C(X)\otimes C(Y))_\bullet},\qquad \EZ\circ \AW\simeq \id_{C_\bullet(X\times Y)}\]

Therefore, combining Lemma 8 with Theorem 9 (Eilenberg-Zilber), we obtain the following result.

Corollary 10 (Künneth) Fix topological spaces \(X,Y\). Then for their product space \(X\times Y\) and a principal ideal domain \(A\), there exists a short exact sequence

\[0 \rightarrow \bigoplus_{p+q=k}H_p(X;A)\otimes_AH_q(Y;A)\rightarrow H_k(X\times Y;A)\rightarrow \bigoplus_{p+q=k-1}\Tor_1^A(H_p(X;A),H_q(Y;A))\rightarrow 0\]

Moreover, this short exact sequence splits (non-canonically), and thus there is an isomorphism

\[H_k(X\times Y;A)\cong \left( \bigoplus_{p+q=k}H_p(X;A)\otimes_AH_q(Y;A)\right)\oplus \left(\bigoplus_{p+q=k-1}\Tor_1^A(H_p(X;A),H_q(Y;A)) \right)\]

Of course, using this result together with Theorem 5 (Universal coefficient theorem for cohomology, general version), one can obtain the cohomology version of the Künneth formula.

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References

[Hat] A. Hatcher, Algebraic Topology. Cambridge University Press, 2022.
[May] J. P. May, A concise course in algebraic topology.

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1. Of course, one would like this pairing to yield a homomorphism from \(H^k(X;A)\) to \(\Hom(H_k(X),A)\), but as we know from Proposition 3 (Universal coefficient theorem for cohomology), the situation is not so simple: an \(\Ext\) term capturing hidden torsion must appear. ↩