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I encountered the statement that $H^*(G,\pi ^*_{S\rightarrow G}A)$ is a universal functor, when S is an open subgroup of $G$. Its statement is accompanied with the reasning that, as $S$ has finite index in $G$, $\pi ^*_{S\rightarrow G}A$ is injective whenever $A$ is. But how to derive this?
P.S. Some additional conditions were missing in this question. I apologize. Here $G$ is a profinite group, which acts on the $G$-module $A$. So $G$ is compact, which justifies the claim that $S$ has finite index in $G$.
Thanks for your attention.

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up vote 1 down vote accepted

The key reason it works is that in this case, $kG$ is (finitely generated) projective as a $kS$-module (i.e. $\pi^*_{S \to G} kG$ is finitely generated projective in your notation). Indeed $kG$ is a free $kS$-module with free generators a set of coset reps for $S$ in $G$. Here $k$ is whatever ring you are working over.

Injectivity of $\pi^*_{S \to G} A$ is equivalent to the vanishing of all groups $\operatorname{Ext}^n_{kS}(-, \pi^*_{S \to G} A)$ with $n>0$. But by the Eckmann-Shapiro Lemma,

$$\operatorname{Ext}^n_{kS}(M, \pi^*_{S \to G} A) \cong \operatorname{Ext}^n_{kG}( M \!\!\uparrow _S ^G, A) = 0 $$

because $A$ is injective.

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So it avails of the theory of finite-dimensional representations to transform the injectivity of A to that of $\pi ^*_{S\rightarrow G}A$. Is this correct? Thanks for the answer. – awllower Dec 6 '12 at 15:47
No, finite-dimensionality isn't needed. – m_t_ Dec 6 '12 at 15:50
As far as I understand, it employs the finitely-generatedness, along with the projectivity of $kG$ to produce an isomorphism relating the extensions over $G$ and $S$. Is this correct, again? Still thanks for the explanation. – awllower Dec 6 '12 at 16:46
Maybe I'm missing something, but I don't actually think any finiteness hypotheses are needed. The ingredients of the proof are first that $kG$ is free over $kS$ and second Eckmann-Shapiro. But you find E-S stated in Benson - Representations and Cohomology I (or in Cartan-Eilenberg) with no assumptions on the rings or modules involved, and $kG$ is always free on a set of coset reps regardless of whether the index is finite. – m_t_ Dec 6 '12 at 18:33
You might like Kenneth Brown's notes on cohomology of groups – m_t_ Dec 8 '12 at 12:09

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