Which conditions do we need to ensure that the cup product on cochains descends to cohomology?

Question

Let $(X_*,\partial)$ be a complete resolution of $G$ and $(M^*,\delta)$ be the corresponding complex by applying Hom functor to the first variable. There exists $G$ homomorphisms $\varphi_{p,q}:X_{p+q}\rightarrow X_p\otimes X_q$ for every $p,q\in \mathbb{Z}$ such that $$\varphi_{p,q} \partial =\partial' \varphi_{p+1,q}+(-1)^p\partial'' \varphi_{p,q+1}\\ (\epsilon\otimes\epsilon) \varphi_{0,0}=\epsilon.$$ Where $\partial'=\partial\otimes 1$ and $\partial''=1\otimes \partial$

Let $A,B$ be $G$ modules and $f\in\rm{Hom}_G(X_p,A)$ and $g\in \rm{Hom}_G(X_q,B)$ we then define cup product

$$f\cup g=(f\otimes g)\varphi_{p,q}$$

Then I need to prove that this induces a well defined map

$$H^p(G,A)\times H^q(G,B)\rightarrow H^{p+q}(G,A\otimes B)$$

I guess it is enought to prove that cup product of two cocyles is a cocycle and cup product of co boundaries is a coboundary.

I see the result $$\delta(f\cup g)=\delta f\cup g+(-1)^pf\cup \delta g$$

Suppose $f,g$ are cocyles then $\delta f=0$ and $\delta g=0$ we then have $\delta(f\cup g)=0$ i.e., $f\cup g$ is a cocycle.

Suppose $f,g$ are coboundaries $f=\delta h$ and $g=\delta h'$ then $\delta(h\cup g)=\delta h\cup g+(-1)^ph\cup \delta(\delta h)=f\cup g$ which says that cup product of coboundaries is coboundary.

I want to know if there is anything else that i have to check for well definedness of cup product.

Answer 1

In order for the cup product to descend to cohomology, one needs to check the following two conditions:

1. A cup product of cocycles is a cocycle.

2. A cup product of a cocycle with a coboundary is a coboundary.

Both follow from the equation you give:

$$\delta(f\cup g)=\delta f\cup g+(-1)^p f\cup\delta g $$

You already checked the first condition. For the second, let $f=\delta g$ be a coboundary and $h$ a cocyle. Then

$$ f\cup h=\delta g \cup h=\delta(g\cup h)$$

where we used that $g$ is a cocycle so that the final term vanishes.

Note that a cup product of a cocycle and coboundary being a coboundary means more than a cup product of coboundaries being one. You need this condition, and not the one you mention, because you want to ensure that you cannot change the cup product by adding a coboundary to one of the two cohomology class representatives that you use to take the cup product. It is clear why we need the first condition.