# Sheafification of a given presheaf

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!

• $\mathcal{F}$ 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. – Roland Mar 19 at 15:30
• @Roland I fixed the statement: the support is not necessarily compact, it is just bounded. – bellcircle Mar 20 at 1:34

## 1 Answer

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 $$[(f,W)]$$.

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