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in the proof of Proposition II, 3.3.7 (ii) -> (iv) in Demazure-Gabriel "Introduction to Algebraic Geometry and Algebraic Groups", it is stated as obvious that an exact sequence of kG-modules $0 \to V' \to V \to V'' \to 0$ induces an exact sequence of Hom-sets, i.e.

$$ 0 \to \text{Hom}(V'', V') \to \text{Hom}(V'', V) \to \text{Hom}(V'', V'') \to 0$$

Whilst I see that this sequence should be left-exact (basically by the same arguments as with usual modules), I do not understand why the last map is surjective. Is the reason connected to the special structure of the last Hom, i.e. that it is $V'' \to V''$?

Remark: I do not know that the original sequence is split, then the statement would be quite obvious.

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  • $\begingroup$ I don't have access to the book, but if $kG$ is semisimple, its higher $Ext$-functors vanish for all modules. This is the case, for instance, if $G$ is finite and of order coprime to the characteristic of $k$. $\endgroup$
    – Sergey Guminov
    Commented Feb 9, 2023 at 21:07
  • $\begingroup$ Actually, I want to show that $kG$ is semi-simple using this argument :) Using this, I want to show that the original sequence of modules splits. $\endgroup$
    – max_121
    Commented Feb 9, 2023 at 21:14
  • $\begingroup$ A short exact sequence of vector spaces splits automatically, albeit highly non-canonically. Can you find a splitting as a $kG$-module? Perhaps using a Reynolds operator argument... $\endgroup$
    – Tabes Bridges
    Commented Feb 9, 2023 at 23:32
  • $\begingroup$ @max_121 - you should include some hypotheses. What is $k$? a field? $\endgroup$
    – hm2020
    Commented Jan 19 at 11:39

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Since the hom-sets are taken over $k$, then indeed it is obvious, as $\hom_k(V'',-)$ is an exact functor. If they are claiming this is true for any $k$ and any $G$, then this is most likely the situation. Presumably the authors write $\hom_G$ for $G$-equivariant hom-sets.

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