Cap product of infinite cyclic group in degree one

Question

I have a question about the cap-product in group cohomology. Maybe someone can help me out or clarify a mistake in my mind.
Suppose we have an infinite cyclic group $G$ with fixed generator $g$. Then obviously $H_1(G,\mathbb{Z})=\mathbb{Z}$ and we can choose a generator $c\in H_1(G,\mathbb{Z})$ (which is unique up to sign). Then consider a $G$-module $M$. Using the cap-product with $c$ we can define a homomorphism of groups
$H^1(G,M)\rightarrow H_0(G,M)$.
My question/hope is now the following: Since $G$ is infinite cyclic generated by $g$, we can view $H^1(G,M)=H_0(G,M)$. I want to work with cochains, i.e. an element in $H^1(G,M)$ is a cocycle $J$ modulo coboundary. In this language, the isomorphism can be given by mapping a cocycle $J$ to the image $J(g)\in M$ (if I am correct). Is it then true, that the cap-product is also given by this map? Also, the generator $c$ is not unique, but the map $J\mapsto J(g)$ also depends on $g$. So is there a 'canocical' choice for $c$?
Thanks for any help.
Edit: Maybe we can choose $c$ 'canocical' via $H_1(G,\mathbb{Z})=G$ and $c$ should correspond to the generator $g$?

Answer 1

There is a canonical/functorial isomorphism between $H_1(G,\mathbb{Z})$ and the abelianization $G_{ab}=G/[G,G]$ which is given by standard resolutions(see e.g. II.3. of Brown's cohomology of groups). This isomorphism sends every element of $G$ precisely to that in $G_{ab}$ as required.

The assertion that $H^1(\mathbb{Z},M)=H_0(\mathbb{Z},M)$ in the 3rd paragraph might also come from standard resolution (I don't know how you deduce this identity, at least I would just use standard resolution, see also III.1.Example.1 of Brown's book). And in this way they are both $M/(g-1)M$. So no matter how $c$ is chosen, it's always specified by $g$(or you can say $c$ is not defined without specifying $g$).

For your second question, yes, the cap product is also given by this evaluation map, this is exactly V.3.10 of Brown's book asserting that $v\cap z=v(z)\in H_0(G,M)=M_G$ for $v\in H^p(G,M), z\in H_p(G,\mathbb{Z})$.