Sheafification of a subpresheaf

Question

Let $\mathscr{F}$ be a subpresheaf of a sheaf $\mathscr{H}$. If I understood correctly, in this case the sheafification has a particularly simple form. It is the sheaf defined by $$\widetilde{\mathscr{F}}(U)=\{s\in \mathscr{H}(U)\:|\:s\text{ locally lies in }\mathscr{F}\},$$ where we say that $s\in \mathscr{H}(U)$ locally lies in $\mathscr{F}$ is there's an open cover $\{U_i\}$ of $U$ such that $s|_{U_i}\in \mathscr{F}(U_i)$ for every $i$.

It is clear that this is indeed a sheaf but I can't see how it satisfies the universal property of the sheafification.

I thought about doing it in the following way: I really think that this construction is functorial in $\mathscr{F}$. Supposing that, given a morphism $\mathscr{F}\to\mathscr{G}$ we should have an induced morphism $\widetilde{\mathscr{F}}\to\widetilde{\mathscr{G}}$ which coincides with our original morphism when restricting to $\mathscr{F}$. If $\mathscr{G}$ is a sheaf, $\widetilde{\mathscr{G}}=\mathscr{G}$ and so this yields our desired morphism. But I can't really see why this construction is functorial in $\mathscr{F}$.

Answer 1

Given a sheaf $\mathscr{G}$ and a morphism of presheaves $f:\mathscr{F}\to \mathscr{G}$, you can directly construct a morphism of presheaves (and therefore of sheavs) $\widetilde{f}:\widetilde{\mathscr{F}}\to \mathscr{G}$ that extends $f$. Given a section $s\in \widetilde{\mathscr{F}}(U)$, choose any open cover $\{U_i\}$ of $U$ such that the restrictions $s\mid_{U_i}$ lie in $\mathscr{F}$. Then you can apply $f$ to each of these restrictions, and glue their images to get $\widetilde{f}(s)\in \mathscr{G}(U)$. It is not hard to check that this is well-defined, i.e. independent of the open cover, and that in fact $\widetilde{f}$ is the unique extension of $f$. The latter shows that $\widetilde{\mathscr{F}}$ is a sheafification.

Regarding the functoriality you mention, since your definition requires choice of a sheaf $\mathscr{H}$, I am not sure how to make sense of this (of course you know after the fact that sheafification is functorial by the universal property).

Answer 2

I am assuming $\mathscr H$ is a sheaf of abelian groups on a topological space $X$.
Since $\mathscr H$ is a sheaf, by the universal property of sheafi-fication, the morphism of pre-sheaves $\iota:\mathscr F\hookrightarrow\tilde {\mathscr F} $ factors through $\mathscr F\xrightarrow{j}\mathscr F^{sh}\xrightarrow{\bar \iota }\tilde{\mathscr F}$. To see that $\bar \iota: \mathscr F^{sh}\rightarrow \tilde {\mathscr F}$ is an isomorphism, it is sufficient to check at the level of stalks.
Observe that at the level of stalks, we have for $x\in X$, the base space, the composition $$\mathscr F_x\xrightarrow{j_x}\mathscr F^{sh}_x\xrightarrow{\bar \iota_x }{\tilde{\mathscr F}}_x=\mathscr F_x $$ which is identity. Moreover, $j_x$ is an isomorphism. This comes from the very construction of the sheafification. Thus $\bar \iota_x$ is an isomorphism for every $x\in X$ and hence $\bar \iota $ is an isomorphism.

Note: The standard construction of $\mathscr F^{sh}$ is to consider $\mathscr F$ as a sub-presheaf of the sheaf of stalks $\mathscr H(U):=\{\prod_{x\in U}s_x: s_x\in \mathscr F_x \}$ and then look at $\tilde {\mathscr F}$ in this sheaf.