Jtdylktuy

Def 1: $alpha:G rightarrow G'$, group homomorphism. If $A'$ is a $G'$ module $f:A rightarrow A'$ is a $Bbb Z$ map we call $(alpha, f)$ a compatible pair if $f:A rightarrow _{alpha}A'$ is a $G$-map, where $_{alpha} A'$ is $G$-module given by map $alpha$.

Def 2: There is an induced map
$$(alpha,f)_*: H_n(G,A) rightarrow H_n(G',A')$$

I am confused with how the map is in fact constructed. Rotman outlines as follows:

Let $P_* rightarrow Bbb Z$ be a $G$-projective resolution of $Bbb Z$. $P'_* rightarrow Bbb Z$ be a $G'$ projective resolution of $Bbb Z$. The functor $U:_{Bbb Z G'} Mod rightarrow _{Bbb Z G} Mod$ the change of groups functor (using $alpha$), gives an acyclic $G$-complex, $$UP'_* rightarrow Bbb Z$$
The comparison theorem gives a map
$$ UP'_* rightarrow P_* $$
over $1_{Bbb Z}$. We apply the functor $-otimes_G A$ [Rotman wrote $-otimes_G {}_{alpha} A$ which does not make sense].Then this gives
$$ tau: UP'_* otimes_G A rightarrow P_* otimes_G A$$
$f:A rightarrow {}_{alpha} A'$ induces a chain map
$$f_*: P_* otimes _G A rightarrow P_* otimes_G {}_{alpha} A' $$

Then nothing makes sense for me

$$f_*tau_* : UP'_* otimes _G A rightarrow P_*otimes_G {}_{alpha} A' $$
Considering homology induces
$$H_n(G,A) rightarrow H_n(G',A').$$

How so?