For the sheaf $\mathscr{F}^+$ associated to the presheaf $\mathscr{F}$, how to prove $\mathscr{F}^+_P=\mathscr{F}_P$?

Question

In Hartshorne's book algebraic geometry, he introduced the concept the sheaf $\mathscr{F}^+$ associated to the presheaf $\mathscr{F}$ in proposition 1.2, chapter II. Then he claim that for each point $P$, the stalks $\mathscr{F}_P=\mathscr{F}^+_P$ without proof. I construct an isomorphism, which is

$\mathscr{F}^+_P\to\mathscr{F}_P$, by $\langle U,f\rangle\mapsto f(P)$,

$\mathscr{F}_P\to\mathscr{F}^+_P$, by $\langle U, s\rangle\mapsto \langle U,\hat{s}\rangle$

where the function $\hat{s}:U\to \bigcup_{Q\in U}\mathscr{F}_Q$ by $Q\mapsto s_Q$. Hence we got isomorphism $\mathscr{F}_P\simeq\mathscr{F}^+_P$. However, I was wondering if there is a more simplical and intuitive method to get this conclusion?

Answer 1

I don't know if this will count as a "simpler and more intuitive method", but it does give a proof at a somewhat higher level which might be enlightening. For this proof, we will use the fact that for a point $P \in X$, the stalk functor ${-}_P : \mathbf{Ab}_{\operatorname{pre}}(X) \to \mathbf{Ab}$ defined by $\mathscr{F} \mapsto \mathscr{F}_P$ has a right adjoint, given by $i_{P*} : \mathbf{Ab} \to \mathbf{Ab}_{\operatorname{pre}}(X)$. This is defined where $i_{P*}(G)$ is the presheaf whose sections over $U$ are $G$ if $P \in U$, or 0 if $P \notin U$; and the restriction maps are $\operatorname{id}_G$ if $P \in V \subseteq U$, or the zero map otherwise. Thus, we have $\operatorname{Hom}(\mathscr{F}_P, G) \simeq \operatorname{Hom}(\mathscr{F}, i_{P*}(G))$.

However, we can check that $i_{P*}(G)$ is actually a sheaf. Therefore, by the universal property of the sheafification, we get a canonical isomorphism $\operatorname{Hom}(\mathscr{F}, i_{P*}(G)) \simeq \operatorname{Hom}(\mathscr{F}^+, i_{P*}(G))$. And now applying the adjunction again in reverse, we see $\operatorname{Hom}(\mathscr{F}^+, i_{P*}(G)) \simeq \operatorname{Hom}(\mathscr{F}^+_P, G)$.

Furthermore, we can see that these isomorphisms are functorial in $G$, so we get isomorphisms of functors $\mathbf{Ab}_{\operatorname{pre}}(X) \to \mathbf{Set}$: $$\operatorname{Hom}(\mathscr{F}_P, {-}) \simeq \operatorname{Hom}(\mathscr{F}, i_{P*}({-})) \simeq \operatorname{Hom}(\mathscr{F}^+, i_{P*}({-})) \simeq \operatorname{Hom}(\mathscr{F}^+_P, {-}).$$ Therefore, applying Yoneda's Lemma, we get a canonical isomorphism $\mathscr{F}_P \simeq \mathscr{F}^+_P$.