1
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ $\mathcal{F}$ 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 Mar 19 at 15:30
  • $\begingroup$ @Roland I fixed the statement: the support is not necessarily compact, it is just bounded. $\endgroup$ – bellcircle Mar 20 at 1:34
0
$\begingroup$

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 $[(f,W)]$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.