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Let $C^n(G,A)$ be the set of continuous functions $G^n \rightarrow A$ with $G$ a profinite group and $A$ a discrete $G$-module (these are the functions that are locally constant). I want to prove that $C^n(G,A) = \varinjlim C^n(G/U,A^U)$, where $U$ runs through all open normal subgroups of $G$ and $A^U$ is the submodule fixed by $U$.

I think that the direct system you have to use is the following: let $U \subset V$ be 2 open normal subgroups of $G$. Then we have a canonical projection $p_{UV}:G/U \rightarrow G/V$, which is clearly continuous since both sides have the discrete topology. We also have a canonical inclusion $i_{UV}:A^V \rightarrow A^U$ which is again continuous because of the discrete topology on both sides. So we can make a function $\rho_{VU}:C^n(G/V,A^V) \rightarrow C^n(G/U,A^U)$, defined by $\rho_{VU}(\phi)=i_{UV}\phi p_{UV}$.

Now, I was trying to make an isomorphism from $C^n(G,A)$ to $\varinjlim C^n(G/U,A^U)$, but I don't see how to do this. I must admit, I'm not very good with limits in categories.

Any help would be appreciated.

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up vote 2 down vote accepted

This looks like it might be some work.

First of all we don't only have maps $C^n(G/V,A^V) \to C^n(G/U,A^U)$ you also have maps $C^n(G/V,A^V) \to C^n(G,A)$ and cannonical maps $C^n(G/V,A^V) \to \varinjlim C^n(G/V,A^V)$.

Putting it all together you have a diagram like this:

Image

where $\Psi$ is induced by the universal property, and everything commutes. Your task now is to show that $\Psi$ is an isomorphism. Unfortunately, I don't see this as an easy task! It seems you need to go through the whole verification that $\Psi$ is both injective and surjective.

Since it looks like a bit of work, I offer up an alternative. If you have institutional access to SpringerLink you should be (at least I can) able to view the book 'Profinite Groups' by Ribes and Zalesskii. Theorem 5.1.4(a) is a proof of a very similar statement - in essence it appears that following their steps should work here (with suitable modifications).

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Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $\Psi$ is injective or surjective. –  KevinDL Apr 2 '12 at 11:40
    
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time! –  Juan S Apr 2 '12 at 11:53
    
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof. –  KevinDL Apr 2 '12 at 11:57
    
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard –  Juan S Apr 2 '12 at 12:01
    
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13. –  KevinDL Apr 2 '12 at 12:06
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