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I have some problems to show that the following contruction 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 $s_x$ un $x$. If with $U$ we indicate a generic open set of $X$, then the set of sections of $\pi$ on $U$ is $$\mathcal F^+(U)=\{\widetilde s:U\rightarrow Et(\mathcal F)\;\;with\;\; \widetilde s(x)=s_x\;\forall s\in \mathcal F(U)\}$$ Whe give a certain topology on $Et(\mathcal F)$ and make $\pi$ and $\widetilde s$ continuous functions. In this way whe 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 exists a unique morphism $\theta:\mathcal F^+\rightarrow\mathcal G$ such that $\theta\circ\phi=\psi?$

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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. –  Zhen Lin May 19 '12 at 10:46
    
a natural way would be to define $\theta(\widetilde s)=\psi(s)$ but what about the uniqueness? –  Galoisfan May 19 '12 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 $\theta = \theta'$. –  Zhen Lin May 19 '12 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. –  Galoisfan May 19 '12 at 14:20
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Crossposted to MO here. –  Zhen Lin May 19 '12 at 19:22
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1 Answer

up vote 5 down vote accepted

Definition 1 Let $\mathcal F$ be a presheaf of sets(or abelian groups or rings, etc.) on a toplogical 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 folowing lemma is fundametal.

Lemma 5 Let $\mathcal F$ be a sheaf on a toplogical 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 $\iota(s)$ = $\sigma$. Hence $\iota$ 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 toplogical 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 toplogical 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: There exists 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.

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