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Let $X$ be a topological space and $U$ be open in $X$. Let $\mathcal F$ be a presheaf of rings on $X$. Let $\mathcal F_u$ denote the presheaf restricted to the open set $U$. $\mathcal F^+$ denote the sheafication of $\mathcal F$. I want to show that $(\mathcal F_u)^+$ is $(\mathcal F^+)_u$ using the universal property of sheafication. Now there is an obvious map $(\mathcal F_u)^+$ to $(\mathcal F^+)_u$. What to get an inverse map and prove they are isomorphic?

Mohan
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2 Answers2

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Let $j : U \to X$ denote the open inclusion. The restriction functor $\mathrm{Sh}(X) \to \mathrm{Sh}(U), F \mapsto F|_U$ is isomorphic to $j^*$ and therefore left adjoint to the direct image functor $j_*$. But these functors are also defined on presheaves with the same formulas and there also satisfy this adjunction. It follows, that for every sheaf $G$ on $U$, we have

$\hom((F|_U)^+,G) = \hom(F|_U,G) = \hom(F,j_* G) = \hom(F^+,j_* G) = \hom((F^+)|_U,G).$

The Yoneda Lemma does the rest.

Martin Brandenburg
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Why not just compute the stalks and see that they are isomorphic? Let $i : U \hookrightarrow X$ denote the inclusion map and let $\mathcal{F}|_{U} := i^{-1}\mathcal{F}$ be the restriction of the presheaf $\mathcal{F}$ to $U$. From the canonical map $\theta : \mathcal{F} \to \mathcal{F}^+$ we get by applying the restriction functor a map (which for simplicity I will just call $\varphi$) $\varphi : \mathcal{F}|_U \to (\mathcal{F}^+)|_U$. The codomain is a sheaf and so the universal property of the sheafification gives a map $\overline{\varphi} : (\mathcal{F}|_U)^+ \to (\mathcal{F}^+)|_U$ such that the diagram

$\hspace{2.in}$ enter image description here

commutes. Now for any $p\in U$ let us pass to the stalk at $p$. The functored vertical column we know is an isomorphism, so we just need to prove that $\varphi_p$ is an isomorphism. To do this, for every $V \subseteq U$ open we have a square

$\hspace{2.5in}$enter image description here

where now I consider $\mathcal{F}$ as a presheaf on $U$. Then when we pass to the direct limit over all open sets $V \subseteq U$ that contain $p$, the functored left column is an isomorphism, and the functored top and bottom rows are isomorphisms as well. Thus $\varphi_p$ is an isomorphism for all $p$ and so $\overline{\varphi}$ is an isomorphism.

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    There seems to be a misconception: a sheaf morphism whose stalks are isomorphisms is itself an isomorphism, but two sheaves whose stalks are isomorphic need not be isomorphic. – Zhen Lin Jul 08 '13 at 11:40
  • Dear @ZhenLin yes indeed a misconception has arisen. In any case, can my answer be rectified? –  Jul 08 '13 at 11:47
  • Well, there is a canonical morphism $(\mathscr{F}|_U)^+ \to (\mathscr{F}^+)|_U$, by the universal property of $(\mathscr{F}|_U)^+$ and the fact that $(\mathscr{F}^+)|_U$ is a sheaf. – Zhen Lin Jul 08 '13 at 11:54
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    If you apply the restriction functor to the sheafification map $F\rightarrow F^+$ (thought of as a map of presheaves) you get a map $F\vert_U\rightarrow(F^+)\vert_U$. The target is actually a sheaf, so this canonically gives rise to a map $(F\vert_U)^+\rightarrow(F^+)\vert_U$. If you can verify that your identifications of the stalks is in fact the map induced by this canonical morphism, you'd win. The Stacks Project does this all the time. – Keenan Kidwell Jul 08 '13 at 11:54
  • And how do you prove that the associated sheaf doesn't change the stalks, without using explicit constructions? Again you have to use adjoint functors! Also note that my proof is valid for any site (in the sense of Grothendieck), but the stalks approach is more indirect and only works for sites with enough points. – Martin Brandenburg Jul 08 '13 at 13:12
  • Dear @KeenanKidwell, I have edited my answer. Though, I am a little unsure about the last part of my proof showing that $\varphi_p$ is an isomorphism for all $p$. –  Jul 08 '13 at 15:04
  • As I said, you use that $F \to F^+$ induces an isomorphism on stalks. Try to derive this from the universal property. You will end up with adjoint functors. – Martin Brandenburg Jul 08 '13 at 18:33
  • Dear @BenjaLim, Yes, your argument looks correct to me. The point I think Martin is trying to make is that you are arguing with stalks, as opposed to with adjoint functors. The former doesn't generalize to arbitrary sites because sheaves on arbitrary sites aren't always determined by their stalks (in the sense to which you're accustomed for sheaves on topological spaces). In fact it's not even immediately clear what the stalk of a sheaf should be in the general setting because the notion of point is not so obvious. But your argument is perfectly correct because, for topological spaces, you – Keenan Kidwell Jul 09 '13 at 22:48
  • have enough points and you can argue with stalks. – Keenan Kidwell Jul 09 '13 at 22:51
  • It's not the only point I make ... – Martin Brandenburg Jul 10 '13 at 08:06