# Sheafification of a presheaf through the etale space

I have some problems to show that the following construction defines a sheafification:

Let $$\mathcal F$$ be a presheaf on $$X$$, and let $$Et(\mathcal F)$$ be the etale space associated to $$\mathcal F$$, with $$\pi:Et(\mathcal F)\rightarrow X$$ that is the canonical map which sends a germ $$s_x$$ to $$x$$. If with $$U$$ we indicate a generic open set of $$X$$, then the set of sections of $$\pi$$ on $$U$$ is $$\mathcal F^+(U)=\{\,\widetilde s:U\rightarrow Et(\mathcal F)\ \ \text{with}\ \ \widetilde s(x)=s_x\;\forall s\in \mathcal F(U)\,\}$$ We give a certain topology on $$Et(\mathcal F)$$ and make $$\pi$$ and $$\widetilde s$$ continuous functions. In this way we define the sheaf $$\mathcal F^+$$ of continuous sections of $$\pi$$, and the morphism (for all $$U$$) $$\begin{eqnarray} \phi(U):\mathcal F(U)&\rightarrow& \mathcal F^+(U)\\ s&\mapsto& \widetilde s \end{eqnarray}$$ Now if $$\mathcal F^+$$ satisfies the "universal property", it is the sheafification of $$\mathcal F$$. Suppose that $$\psi$$ is a morphism from $$\mathcal F$$ in a generic sheaf $$\mathcal G$$; how can I prove that there exists a unique morphism $$\theta:\mathcal F^+\rightarrow\mathcal G$$ such that $$\theta\circ\phi=\psi?$$

• One way to do it is to show that there is a natural bijection between sheaf morphisms $\mathscr{G} \to \mathscr{H}$ and bundle morphisms $\textrm{Ét}(\mathscr{G}) \to \textrm{Ét}(\mathscr{H})$, or equivalently, that there is an isomorphism from the sheaf of sections of $\textrm{Ét}(\mathscr{G})$ to the sheaf $\mathscr{G}$ itself. Commented May 19, 2012 at 10:46
• a natural way would be to define $\theta(\widetilde s)=\psi(s)$ but what about the uniqueness? Commented May 19, 2012 at 10:56
• That formula does not obviously define $\theta$ – you have to check that it makes sense. As usual, uniqueness is actually the easy part: once you convince yourself that there is a $\theta$, notice that $\phi$ is surjective on stalks, so if $\theta$ and $\theta'$ are sheaf morphisms, then $\theta \circ \phi = \theta' \circ \phi$ implies $\theta = \theta'$. Commented May 19, 2012 at 12:36
• You're right, $\theta(\widetilde s)=\psi(s)$ doesn't define a function because if $s_x=t_x$ for all $x$, doesn't imply that $s=t$. This because $\mathcal F$ is a presheaf. Commented May 19, 2012 at 14:20
• Crossposted to MO here. Commented May 19, 2012 at 19:22

Definition 1. Let $$\mathcal F$$ be a presheaf of sets (or abelian groups or rings, etc.) on a topological space $$X$$. Let $$Et(\mathcal F)$$ be the disjoint union $$\cup_{x \in X} \mathcal F_x$$. Let $$U$$ be an open subset of X. Let $$s \in \mathcal F(U)$$. We denote by $$[U, s]$$ the subset {$$s_x; x \in U$$} of $$Et(\mathcal F)$$. Let Open$$(X)$$ be the set of open subsets of $$X$$. We define a topology on $$Et(\mathcal F)$$ as the one generated by the subset {$$[U, s]; U \in$$ Open($$X), s \in \mathcal F(U)$$} of the power set of $$Et(\mathcal F)$$. Let $$\pi:Et(\mathcal F) \rightarrow X$$ be the canonical map which sends a germ $$s_x$$ to $$x$$. Let $$\mathcal F^+(U)$$ be the set {$$f:U \rightarrow Et(\mathcal F)$$; $$f$$ is a continuous map and $$\pi f = id_U$$}. Clearly $$\mathcal F^+(U)$$ defines a sheaf $$\mathcal F^+$$ on $$X$$.

Lemma 2. Let $$\mathcal F^+(U)$$ be as above. Then $$\mathcal F^+(U)$$ = {$$f:U \rightarrow Et(\mathcal F)$$; $$f$$ is a map such that for each $$x \in U$$ there exists an open neighborhood $$U_x$$ of $$x$$ contained in $$U$$ and $$s \in \mathcal F(U_x)$$ such that $$f(y) = s_y$$ for each $$y \in U_x$$}.

Proof: Clear.

Definition 3. Let $$\mathcal F$$ be a presheaf on a toplogical space $$X$$. Let $$U$$ be an open subest of $$X$$. Let $$s \in \mathcal F(U)$$ We define a map $$\tilde{s}: U \rightarrow Et(F)$$ by $$\tilde{s}(x)$$ = $$s_x$$ for each $$x \in U$$. Clearly $$\tilde{s} \in \mathcal{F^+(U)}$$.

Definition 4. Let $$\mathcal F$$ be a presheaf on a toplogical space $$X$$. Let $$U$$ be an open subest of $$X$$. We define a map $$\iota_U: \mathcal F(U) \rightarrow \mathcal F^+(U)$$ by $$\iota_U(s) = \tilde{s}$$, where $$\tilde{s}$$ is defined in Definition 3. Clearly $$\iota_U$$'s define a morphism $$\iota:\mathcal F \rightarrow \mathcal F^+$$. We call $$\iota$$ the canonical morphism.

The following lemma is fundamental.

Lemma 5. Let $$\mathcal F$$ be a sheaf on a topological space $$X$$. Then the canonical morphism $$\iota:\mathcal F \rightarrow \mathcal F^+$$ is an isomorphism.

Proof: Let $$U$$ be an open subest of $$X$$. It suffices to prove that $$\iota_U: \mathcal F(U) \rightarrow \mathcal F^+(U)$$ is an isomorphism. Let $$s$$ and $$t$$ be $$\in \mathcal F(U)$$. Suppose $$\iota_U(s)$$ = $$\iota_U(t)$$. This means that $$s_x$$ = $$t_x$$ for each $$x \in U$$. Hence there exists an open neghborhood $$U_x$$ of $$x$$ for exch $$x \in U$$ such that $$s|U_x$$ = $$t|U_x$$. Since $$\mathcal F$$ is a sheaf, $$s$$ = $$t$$. Hence $$\iota$$ is injective.

It remains to prove that $$\iota$$ is surjective. Let $$\sigma \in \mathcal F^+(U)$$. There exists an open cover $$U_i$$ of $$U$$ and $$s_i \in \mathcal F(U_i)$$ such that $$\sigma(x) = s_i(x)$$ for each $$x \in U_i$$. Since $$s_i|U_i \cap U_j = s_j|U_i \cap U_j$$ by the above claim, there exists $$s \in \mathcal F(U)$$ such that $$s|U_i$$ = $$s_i$$ for each $$i$$. Hence $$\iota(s)$$ = $$\sigma$$, so $$\iota$$ is surjective. QED

Lemma 6. Let $$\mathcal F$$ be a presheaf on a toplogical space $$X$$. Then $$\mathcal F^+_x$$ = $$\mathcal F_x$$ for each $$x \in X$$.

Proof: Clear.

Lemma 7. Let $$\mathcal F$$ and $$\mathcal G$$ be presheaves on a topological space $$X$$. Let $$f:\mathcal F \rightarrow G$$ be a morphism. Let $$U$$ be an open subset of $$X$$. Let $$\sigma \in \mathcal F^+(U)$$. Then the map $$f^+_U(\sigma):U \rightarrow G^+(U)$$ which sends $$x \in U$$ to $$f_x(\sigma(x))$$ for each $$x \in U$$ belongs to $$\mathcal G^+(U)$$.

Proof: Clear.

Lemma 8. Let $$\mathcal F$$ and $$\mathcal G$$ be presheaves on a topological space $$X$$. Let $$f:\mathcal F \rightarrow G$$ be a morphism. Let $$\iota_{\mathcal F}:\mathcal F \rightarrow \mathcal F^+$$ and $$\iota_{\mathcal G}:\mathcal G \rightarrow \mathcal G^+$$ be the canonical morphisms. Then there exists a unique morphism $$f^+:\mathcal F^+ \rightarrow \mathcal G^+$$ such that $$f^+\iota_{\mathcal F} = \iota_{\mathcal G} f$$.

Proof: We have the canonical morphism $$f_x:\mathcal F_x \rightarrow \mathcal G_x$$ for each $$x \in X$$. Let $$U$$ be an open subset of $$X$$. We define a map $$f^+_U:\mathcal F^+(U) \rightarrow \mathcal G^+(U)$$ by sending $$\sigma \in \mathcal F^+(U)$$ to $$f^+_U(\sigma) \in \mathcal G^+(U)$$, where $$f^+_U(\sigma)$$ is defined in Lemma 7. Clearly this gives a morphism of presheaves $$f^+:\mathcal F^+ \rightarrow \mathcal G^+$$ and $$f^+\iota_{\mathcal F} = \iota_{\mathcal G} f$$.

It remains to prove the uniqueness of $$f^+$$. Let $$\psi:\mathcal F^+ \rightarrow \mathcal G^+$$ be a morphism such that $$\psi\iota_{\mathcal F} = \iota_{\mathcal G} f$$. Since $$\mathcal F^+_x$$ = $$\mathcal F_x$$ by Lemma 6, $$\psi_x$$ = $$f^+_x$$ for each $$x \in X$$. Since $$\mathcal F^+$$ and $$\mathcal G^+$$ are sheaves by Definition 1, $$\psi$$ = $$f^+$$. QED

Proposition. Let $$\mathcal F$$ be a presheaf on a toplogical space $$X$$. Let $$\mathcal G$$ be a sheaf on a toplogical space $$X$$. Let $$f:\mathcal F \rightarrow \mathcal G$$ be a morphism. Then there exists a unique morphism $$\theta:\mathcal F^+ \rightarrow \mathcal G$$ such that $$\theta\iota$$ = $$f$$, where $$\iota:\mathcal F \rightarrow \mathcal F^+$$ is the canonical morphism.

Proof: This follows immediately from Lemma 5 and Lemma 8.

• I am uncertain about the characterization of the topology in Def 1. If our etale space is, say, a trivial line bundle over the real line, then sets like $[U,s]$ don't seem to be open. For example, if we take the diagonal section $d$ and the zero section, then $d^{-1}([U,0])$ is just the origin, so $d$ wouldn't be continuous.
– C.D.
Commented Apr 7, 2020 at 2:19