Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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:


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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.