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I'll ask a specific question first, but I believe my question might have a rather immediate abstraction, with which I'll finish.

Let $H,G$ be finite affine group schemes over an algebraically closed field $k$. Here finite means $k[H],k[G]$ are finite-dimensional. When do injections $H\hookrightarrow G$ induces surjections $H^*(G,k)\twoheadrightarrow H^*(H,k)$ in cohomology? Here I view $H^*(G,k)$ as $\mathrm{Ext}^*_{kG}(k,k)$ where $kG=k[G]^{\#}$

What I mean by "when" is what conditions on $H$ and $G$ (if any) are required? Is this true if $H$ and $G$ are constant group schemes associated to finite groups? What if $H$ and $G$ are the group schemes associated to two restricted Lie algebras $\mathfrak{h}$ and $\mathfrak{g}$ (in the case $\mathrm{char}\;k=p$)?

More generally, suppose that $\mathcal{C}$ is an abelian category, and that $F$ is a cohomological functor from $\mathcal{C}$ to abelian groups. When can we expect $F$ to take monomorphisms to surjections?

I'd be happy with an answer to even the restricted Lie algebra case, but welcome any insight into the larger forces at work here (so long as I can see the connection with what I'm interested in, which is the group scheme questions). Thanks!

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One trivial answer: when every monomorphism splits, e.g. when your abelian category is $\mathbf{Vect}$. – Zhen Lin May 13 '14 at 7:30
I don't see how your "More generally ..." generalizes the original question. Finite affine group schemes don't form an abelian category. – Jeremy Rickard May 13 '14 at 8:47
Even for finite groups (constant finite group schemes) this is non-trivial. Let $H\leq G$, and suppose the index of $H$ in $G$ is prime to $p$. Then $H^*(G,k)\to H^*(H,k)$ is always injective when $k$ has characteristic $p$, and a deep theorem of Mislin gives necessary and sufficient conditions for it also to be surjective: for every $p$-subgroup $Q$ of $H$, the natural homomorphism $N_H(Q)/C_H(Q)\to N_G(Q)/C_G(Q)$ must be an isomorphism. I don't know a general group-theoretic condition that is equivalent to $H^*(G,k)\to H^*(H,k)$ being surjective when $p$ divides the index. – Jeremy Rickard May 13 '14 at 15:04

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