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See this link and the second comment to the question in particular: Sheafs of abelian groups are the same as $\underline{\mathbb{Z}}$-modules

To use the universal property of sheafification, I took the ring action as a map from the ring to the ring of endomorphisms of modules so as to use the universal property to get a map from the constant sheaf to the ring of endomorphisms of the sheaf of abelian groups.

But I can't prove that the endomorphism rings form a presheaf, let alone a sheaf. I can't find the restriction maps. Let $M$ be the sheaf of abelian groups. What would be the restriction map between $End(M(U))$ and $End(M(V))$ ?

Jehu314
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    For a sheaf of abelian groups $M$, there is an endomorphism sheaf of $M$ called $\mathcal{H}om(M,M)$. Over any open set $U$ the sections of this sheaf are the endomorphisms $\mathcal{H}om(M|_U,M|_U)$ of the restriction of the sheaf $M$ to $U$. See e.g. this question. – Eoin May 28 '18 at 17:30
  • Thanks a lot, I get it. – Jehu314 May 29 '18 at 01:30

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