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Let $\mathcal{F}$ be a presheaf on $\mathbb{R}$ such that $\mathcal{F}(U)$ is the abelian group of continuous functions with bounded support on $U$. Then what is the sheafification of $\mathcal{F}$?

I guess the sheafification should be the abelian group of continuous functions, but how should I prove it rigorously?

Thanks in advance!

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    $\begingroup$ $\mathcal{F}$ is not a presheaf : if $s\in\mathcal{F}(\mathbb{R})$ has support in $[0,1]$, then $s|_{(0,1)}$ does not have compact support. $\endgroup$
    – Roland
    Commented Mar 19, 2019 at 15:30
  • $\begingroup$ @Roland I fixed the statement: the support is not necessarily compact, it is just bounded. $\endgroup$
    – bellcircle
    Commented Mar 20, 2019 at 1:34

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Let $F$ be our presheaf of continuous functions with bounded support, $G$ be the sheaf of continuous functions, $F^+$ the sheafification of $F$.

Every continuous function with bounded support is a continuous function, so we have an injective morphism $F\to G$, which induces a map $F^+\to G$ by definition of the sheafification. To show that this is an isomorphism, we just need to prove that it is an isomorphism on stalks. However, $F^+$ has the same stalks as $F$, so it suffices to prove that the map $F\to G$ induces an isomorphism on stalks.

However, this is fairly clear. We already know injectivity, so it just remains to check surjectivity. Pick $x\in\Bbb{R}$. Choose $U$ some bounded open neighborhood of $\Bbb{R}$. Then for any $[(f,W)]\in G_x$, we have that the class $[(f|_{W\cap U},W\cap U)]\in F_x$ maps to $[(f,W)]$.

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  • $\begingroup$ @anonymous_downvoter If there is an error in this answer, I would appreciate a comment so that I could improve my answer. $\endgroup$
    – jgon
    Commented Mar 23, 2019 at 23:42

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