Homology/Cohomology of Free Product I recently completed an exercise showing that 
$$
  H_1(G*H,A) \cong H_1(G,A) \oplus H_1(H,A) 
$$
for $A$ a trivial $G*H$-module, and also proved a similar statement for cohomology. This is exercise 6.2.5 in Weibel's Homological Algebra.
Now I'm going back and seeing if I can prove this in a different way than I originally did. 
Here is my attempt:
Using the fact that
$$
  H_1(G,A) \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A
$$
for $A$ a trivial $G$-module, we have
\begin{align*}
  H_1(G*H,A)
  &\cong \frac{G*H}{[G*H,G*H]} \otimes_{\mathbb Z(G*H)} A \\
  &\cong \left( \frac{G}{[G,G]} \oplus \frac{H}{[H,H]} \right) \otimes_{\mathbb Z(G*H)} A \\
  &\cong \left( \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A  \right) \oplus \left( \frac{H}{[H,H]} \otimes_{\mathbb Z(G*H)} A  \right) .
\end{align*}
Does it follow from some kind of change of base theorem that
$$
  \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A 
  \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A  ?
$$
 A: You implicitly are seeing $G/[G,G]$ as a right $\mathbb Z(G*H)$-module. The action of elements of $G$ is the obvious one, and the action of those of $H$ is trivial. 
Now, the tensor product $G/[G,G]\otimes_{\mathbb Z[G*H]}A$ is the quotient of $G/[G,G]\otimes_{\mathbb Z}A$ by the abelian subgroup generated by the elements of the form $$xg\otimes a-x\otimes ga,\qquad g\in G, x\in G/[G,G], a\in A \tag{$\clubsuit$}$$ and those of the form $$xh\otimes a-x\otimes ha,\qquad h\in H, x\in G/[G,G], a\in A.\tag{$\spadesuit$}$$ This is not obvious, but easy to see. 
Now the elements of listed in ($\spadesuit$) are all zero! It follows that $G/[G,G]\otimes_{\mathbb Z[G*H]}A$ is actually the quotient of 
$G/[G,G]\otimes_{\mathbb Z}A$ by the abelian subgroup generated by the elements listed in ($\clubsuit$), which is $G/[G,G]\otimes_{\mathbb Z[G]}A$.
N.B. The general statement underlying this is the following: if $R$ is a ring and $S\subseteq R$ is a set which generates $R$ as a ring, and $M$ and $N$ are right and left  $R$-modules, respectively, then $M\otimes_RN$ is the quotient of $M\otimes_{\mathbb Z}N$ by the subgroup generated by the elements $$ms\otimes n-m\otimes sn, \qquad m\in M,\quad n\in N,\quad s\in S.$$
A: Here is another way to think about this, using topological ideas. Let $BG$ and $BH$ be the classifying spaces of the groups $G$ and $H$. Then, there is a weak equivalence $BG\lor BH\to B(G*H)$ by van-Kampen. Thus this induces an isomorphism on homology and cohomology groups $H_i(BG\lor BH;A)\xrightarrow{\sim} H_i(B(G*H);A)=H_i(G*H,A)$ and $H^i(G*H;A)\xrightarrow\sim H^i(BG\lor BH; A)$.
Furthermore, by Mayer-Vietoris there are isomorphisms $H^i(BG\lor BH;A)\cong H^i(BG;A)\oplus H^i(BH;A)=H^i(G;A)\oplus H^i(H;A)$ and $H_i(BG\lor BH;A)\cong H_i(G;A)\oplus H_i(H;A)$ for each $i>0$.
