Hatcher defines the cup product for cochains $\phi \in C^k(X;R)$ and $\psi \in C^l(X;R)$ as an element $\phi \smile \psi \in C^{k+l}(X;R)$. This cochain is defined on the simplices as follows $\phi\smile\psi(\sigma)=\phi(\sigma|[v_0,...,v_k])\psi(\sigma|[v_k,..,v_l])$.

Now here I think he implicitly assumes that the chains of simplices i.e $C_i(X)$ are in Z coefficients. But while talking about the Kunneth formula he takes $H^*(X;R)$ to be modules over $R$ where R is a commutative ring suggesting the chain groups are themselves modules over R. Now he does mention in passing that if $F$ is a field the dual complex $Hom_F(C_n(X;F),F)$ is the same as $Hom(C_n(X),F)$ identifying both as functions from singular n-simplices to F. And hence the cohomology with homology in $Z$ coefficients or $F$ coefficients agrees.

So firstly I want to know how the isomorphism map $Hom_F(C_n(X;F),F)\rightarrow Hom(C_n(X),F)$ commutes with the boundary maps in the respective chain complexes to give isomorphic cohomology groups either ways.

Secondly, How does one extend the same argument for when coefficients are rings.(I did see somewhere that if cohomology with homology in Z coefficients and cohomology with homology in R coefficients agree if the coefficients of cohomology are taken in the same R).

Thirdly, if the cup product commutes with this "change of coefficients" map i.e the one that takes chains in $Hom(C_i(X);R)$ to $Hom_R(C_i(X;R),R)$.


1 Answer 1


As you say, there is an ambiguity in the choice of coefficients for the chain complex. If $R$ is a commutative ring, then $Hom_R(C_i(X; R), R)$ and $Hom_{\mathbb{Z}}(C_i(X;\mathbb{Z}), R)$ are canonically isomorphic. This is because both $C_i(X;R)$ and $C_i(X; \mathbb{Z})$ are free modules over their base ring, so in order to specify a map to $R$, it suffices to dictate where the base elements go (the base elements being the $n$-cells $\sigma \colon \Delta^n \to X$).

Since the chain maps and the cup product are all be defined on the level of $n$-cells and linearly extended afterwards, the fact that they are compatible with the isomorphism mentioned above is effectively automatic.


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