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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

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Maybe I'm goofing here and confuse finiteness conditions, but doesn't this follow from the fact that $H^\ast(G,-) = \operatorname{Ext}_{\mathbb{Z}G}^{\ast}(\mathbb{Z},-)$ together with the fact that the $FP$-condition says that $\mathbb{Z}$ is a $\mathbb{Z}G$-module of type $FP$, hence that Ext-functor commutes with direct limits and then write your direct sum as a direct system in the obvious way? This should be discussed in Brown's book. – t.b. Jun 6 '11 at 18:23
Tomorrow I'll take a look at that book and then I'll let you know. Thanks you very much! – fatoddsun Jun 6 '11 at 18:36
Brown has this as an unproven theorem, but he gives a reference. – jd.r Jun 6 '11 at 18:58
up vote 4 down vote accepted

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.

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If I understand correctly, this is the essence of the argument I outlined in my comment above, isn't it? – t.b. Jun 6 '11 at 18:53
I believe the OP was stuck on the idea of trying to use the finite length of the resolution rather than the fact that the modules are finitely generated. Your comment is more general than what I wrote, but it does not inform the OP where the hypothesis is being used. – Robert Bell Jun 6 '11 at 19:17
My comment was by no means meant as criticism if you should have understood it that way! (the vote on your answer is from me, in fact). I simply asked because it is a bit outside of my expertise. Thanks for explaining and yes, I agree, I should have mentioned finite generation explicitly. Good point. – t.b. Jun 6 '11 at 19:21
Well, thanks a lot to both of you! Actually I was focusing on the finite length of the resolution instead of the property of the modules of being finitely generated. Now everything works! Thank you again! – fatoddsun Jun 6 '11 at 21:34

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