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.
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).