Direct sum commuting with homology functor

Question

I'm trying to understand a fact about commutation between homology functors and direct sums. In particular, let $G$ be a group of type $FP$ (i.e. there exists a projective resolution of finite length $P_\bullet\to\mathbb{Z}$ over $\mathbb{Z}$G): I should prove that the functor $H^k(G,-)$ commutes with the direct sum. I don't get how to use the assumption about the type $FP$: I've tried to write a proof but in my mind I cannot relate the commutativity with that property. I mean: I'm trying to explicitly write elements in $H^k(G,\bigoplus_i A_i)$ and in $\bigoplus_i H^k(G,A_i)$ in order to find a nice map between them, but the definition is too complicated because I have to take elements which belong to the kernel but not to the image of some boundary operators in a chain complex obtained applying the functor $Hom_G(-,A_i)$ to another chain complex which is itself a resolution of projective modules etc. etc. well... it's just too much! Could you help me with that, please? Thank you, bye

Answer 1

If $G$ is type $FP$ (or $FP_{\infty}$), then there is a resolution of $\mathbb{Z}$ by finitely generated projective $\mathbb{Z}G$-modules. (That they are finitely generated is the key property that you need.)

Suppose that $f \in Hom_G(P, \bigoplus M_i)$. Then there appears to be an induced element of $\bigoplus Hom(P, M_i)$ defined by $f_i(p) = (\pi_i \circ f)(p)$, where $\pi_i$ is the projection onto the $i$-th factor. The devil is in the details however. An element of $\bigoplus Hom(P, M_i)$ must have that all but finitely many $f_i$ are zero, whereas you are given that for each $p$, $f_i(p)$ is zero for all but finitely many $i$. But if $P$ is finitely generated, you can show that indeed all but finitely many $f_i$ are the zero map.