Presheaves Sheaves Stalks and Sheafification

Apply [L1] to the empty cover of $\mathrm{\varnothing }$. The empty family of local sections is vacuously compatible, so gluing produces at least one section $s\in \mathcal{F}(\mathrm{\varnothing })$.

Let $t\in \mathcal{F}(\mathrm{\varnothing })$. Then $t$ also restricts to the same empty family on the empty cover. By the uniqueness part of [L1], one has $t=s$. Therefore $\mathcal{F}(\mathrm{\varnothing })=\{s\}$.

Assume $\mathcal{F}$ is a sheaf. For any $s\in \mathcal{F}(U)$, the two families $d_0(e(s))$ and $d_1(e(s))$ are equal because both entries on $U_i\cap U_j$ are the common restriction $s{\mathrm{\mid }}_{U_i\cap U_j}$ by [F1]. Thus $d_{0}e=d_{1}e$.

Let $(s_i{)}_{i\in I}\in {\prod }_i\mathcal{F}(U_i)$ satisfy $d_0((s_i))=d_1((s_i))$. Unwinding the definitions, this says exactly that $s_i{\mathrm{\mid }}_{U_i\cap U_j}=s_j{\mathrm{\mid }}_{U_i\cap U_j}$ for all $i,j$. By the gluing clause of [L1], there exists a unique $s\in \mathcal{F}(U)$ with $s{\mathrm{\mid }}_{U_i}=s_i$ for all $i$. Hence every equalizing family factors uniquely through $e$, so [L2] shows that $e$ is an equalizer.

Conversely, assume $e$ is an equalizer for every open set and cover. If $s,t\in \mathcal{F}(U)$ satisfy $s{\mathrm{\mid }}_{U_i}=t{\mathrm{\mid }}_{U_i}$ for all $i$, then $e(s)=e(t)$. Let $\ast$ be a singleton and define $h:\{\ast \}\to {\prod }_i\mathcal{F}(U_i)$ by $h(\ast )=e(s)=e(t)$. Then $d_{0}h=d_{1}h$, so [L2] gives a unique $u:\{\ast \}\to \mathcal{F}(U)$ with $eu=h$. The two maps sending $\ast$ to $s$ and to $t$ both satisfy this condition, hence are equal and therefore $s=t$. So locality holds. If $(s_i)$ is a compatible family on the cover, then $d_0((s_i))=d_1((s_i))$, so [L2] yields a unique $s\in \mathcal{F}(U)$ with $e(s)=(s_i)$. That is exactly unique gluing. Therefore [L1] implies that $\mathcal{F}$ is a sheaf.

Assume $\mathcal{F}$ is a sheaf. Let $U={\bigcup }_i{B}_i$ be a basis cover and let $(s_i)$ satisfy the basis-overlap hypothesis. Fix $i,j$. The sets $C\in \mathcal{B}$ with $C\subseteq B_i\cap B_j$ cover $B_i\cap B_j$ by [L1] and the intersection hypothesis on $\mathcal{B}$. On each such $C$ the restrictions of $s_i$ and $s_j$ agree, so locality from [L2] gives $s_i{\mathrm{\mid }}_{B_i\cap B_j}=s_j{\mathrm{\mid }}_{B_i\cap B_j}$. Gluing in [L2] then produces a unique $s\in \mathcal{F}(U)$ with $s{\mathrm{\mid }}_{B_i}=s_i$.

Assume the displayed basis condition. Let $U={\bigcup }_{\alpha \in A}{U}_{\alpha }$ be an arbitrary open cover and let $t_{\alpha }\in \mathcal{F}(U_{\alpha })$ be compatible on overlaps. For each $x\in U$, choose $\alpha (x)$ with $x\in U_{\alpha (x)}$, and then choose $B_x\in \mathcal{B}$ with $x\in B_x\subseteq U_{\alpha (x)}$ by [L1]. Put $r_x:=t_{\alpha (x)}{\mathrm{\mid }}_{B_x}$.

Let $C\in \mathcal{B}$ satisfy $C\subseteq B_x\cap B_y$. Then $C\subseteq U_{\alpha (x)}\cap U_{\alpha (y)}$, so compatibility of the original family gives $r_x{\mathrm{\mid }}_C=t_{\alpha (x)}{\mathrm{\mid }}_C=t_{\alpha (y)}{\mathrm{\mid }}_C=r_y{\mathrm{\mid }}_C$. Therefore the family $(r_x{)}_{x\in U}$ satisfies the displayed basis condition for the basis cover $U={\bigcup }_{x\in U}{B}_x$.

By the assumed basis condition applied to the compatible family from step 2.1, there is a unique $s\in \mathcal{F}(U)$ with $s{\mathrm{\mid }}_{B_x}=r_x$ for every $x\in U$.

Fix $\alpha \in A$. For each $y\in U_{\alpha }$, choose $C_y\in \mathcal{B}$ with $y\in C_y\subseteq B_y\cap U_{\alpha }$ by [L1] and the basis-refinement hypothesis. Then the $C_y$ form a basis cover of $U_{\alpha }$. Since $C_y\subseteq U_{\alpha (y)}\cap U_{\alpha }$, step 2.1 gives $r_y{\mathrm{\mid }}_{C_y}=t_{\alpha (y)}{\mathrm{\mid }}_{C_y}=t_{\alpha }{\mathrm{\mid }}_{C_y}\mathrm{.}$ Because $s{\mathrm{\mid }}_{C_y}=r_y{\mathrm{\mid }}_{C_y}$ by step 3.1, the two sections $s{\mathrm{\mid }}_{U_{\alpha }}$ and $t_{\alpha }$ restrict to the same family on the basis cover $\{C_y{\}}_{y\in U_{\alpha }}$. By the uniqueness part of the assumed basis condition, $s{\mathrm{\mid }}_{U_{\alpha }}=t_{\alpha }$. Since this holds for every $\alpha$, the arbitrary compatible family glues uniquely, so [L2] implies that $\mathcal{F}$ is a sheaf.

If $\mathcal{F}$ is a sheaf of groups, rings, or left $R$-modules, then [F1] already says that its underlying set-valued presheaf is a sheaf.

Conversely, assume the underlying set-valued presheaf is a sheaf. The sets $\mathcal{F}(U)$ already carry the given group, ring, or left $R$-module structures, and the restriction maps already preserve those structures by [F1]. Since [L1] tests only locality and gluing of the underlying sections, the assumed setwise sheaf condition is exactly the required algebra-valued sheaf condition.

Reflexivity holds because if $(U,s)$ is any pair, then $s{\mathrm{\mid }}_U=s{\mathrm{\mid }}_U$, so $(U,s){\sim }_x(U,s)$. Symmetry holds because $s{\mathrm{\mid }}_W=t{\mathrm{\mid }}_W$ implies $t{\mathrm{\mid }}_W=s{\mathrm{\mid }}_W$ on the same neighbourhood $W$.

Suppose $(U,s){\sim }_x(V,t)$ and $(V,t){\sim }_x(Z,r)$. Choose neighbourhoods $W_1\subseteq U\cap V$ and $W_2\subseteq V\cap Z$ with $s{\mathrm{\mid }}_{W_1}=t{\mathrm{\mid }}_{W_1}$ and $t{\mathrm{\mid }}_{W_2}=r{\mathrm{\mid }}_{W_2}$. Then $W:=W_1\cap W_2$ is an open neighbourhood of $x$ contained in $U\cap Z$, and on $W$ one has $s{\mathrm{\mid }}_W=t{\mathrm{\mid }}_W=r{\mathrm{\mid }}_W$. Hence $(U,s){\sim }_x(Z,r)$.

Steps 1.1 and 1.2 show that ${\sim }_x$ is reflexive, symmetric, and transitive. By [F2], it is an equivalence relation.

If $s=0$ in $\mathcal{F}(U)$, then for every $x\in U$ the induced germ is $s_x=0_x$ by the definition of $0_x$ in [F2].

Assume $s_x=0$ for every $x\in U$. Fix $x\in U$. By lem-germ-equivalence-relation, equality of the two germ classes means that their representatives agree on some open neighbourhood $V_x\subseteq U$ of $x$, so $s{\mathrm{\mid }}_{V_x}=0{\mathrm{\mid }}_{V_x}$. The sets $V_x$ cover $U$.

On the open cover $\{V_x{\}}_{x\in U}$, the sections $s$ and $0$ have the same restriction by step 1.2. Locality in [L1] therefore gives $s=0$ in $\mathcal{F}(U)$. For $U=\mathrm{\varnothing }$, this uses the empty cover, which is included in [L1]. Together with step 1.1, this proves both directions.

Fix an open set $U\subseteq X$ and a section $s\in \mathcal{F}(U)$. For every $x\in U$, the hypothesis ${\phi }_x={\psi }_x$ and [F2] give $({\phi }_U(s){)}_x={\phi }_x(s_x)={\psi }_x(s_x)=({\psi }_U(s){)}_x\mathrm{.}$

By [F3], equality of the germs in step 1.1 means that for each $x\in U$ there exists an open neighbourhood $V_x\subseteq U$ such that ${\phi }_U(s){\mathrm{\mid }}_{V_x}={\psi }_U(s){\mathrm{\mid }}_{V_x}\mathrm{.}$ The sets $V_x$ cover $U$.

By [L1], the two sections ${\phi }_U(s)$ and ${\psi }_U(s)$ are equal on $U$. Since $U$ and $s$ were arbitrary, the component maps ${\phi }_U$ and ${\psi }_U$ agree for every open $U$. Therefore $\phi =\psi$ by [F1].

If $\phi$ is an isomorphism with inverse $\psi$, then for every $x$ the induced maps ${\phi }_x$ and ${\psi }_x$ are inverse because taking stalks preserves composition and identities by [F2]. Hence each ${\phi }_x$ is a bijection.

Assume now that every ${\phi }_x$ is a bijection. Fix an open set $U\subseteq X$ and a section $t\in \mathcal{G}(U)$. For each $x\in U$, choose $a_x\in {\mathcal{F}}_x$ with ${\phi }_x(a_x)=t_x$. Pick a representative $s_x\in \mathcal{F}(V_x)$ of $a_x$ on some open neighbourhood $V_x\subseteq U$ of $x$. Then $({\phi }_{V_x}(s_x){)}_x=t_x$, so [F3] lets us shrink $V_x$ and assume ${\phi }_{V_x}(s_x)=t{\mathrm{\mid }}_{V_x}$.

If $s,s^{\mathrm{\prime }}\in \mathcal{F}(U)$ satisfy ${\phi }_U(s)={\phi }_U(s^{\mathrm{\prime }})$, then for every $x\in U$ one has ${\phi }_x(s_x)={\phi }_x(s_x^{\mathrm{\prime }})$ by [F2]. Injectivity of ${\phi }_x$ gives $s_x=s_x^{\mathrm{\prime }}$. By [F3], for each $x$ there exists an open neighbourhood $W_x\subseteq U$ with $s{\mathrm{\mid }}_{W_x}=s^{\mathrm{\prime }}{\mathrm{\mid }}_{W_x}$. The sets $W_x$ cover $U$, so [L2] gives $s=s^{\mathrm{\prime }}$. Thus each ${\phi }_U$ is injective.

Let $x,y\in U$. For any $z\in V_x\cap V_y$, the two sections $s_x{\mathrm{\mid }}_{V_x\cap V_y}$ and $s_y{\mathrm{\mid }}_{V_x\cap V_y}$ have images under $\phi$ whose germs at $z$ both equal $t_z$. Since ${\phi }_z$ is injective, the germs of $s_x$ and $s_y$ at $z$ are equal. By [F3], there is a neighbourhood $W_z\subseteq V_x\cap V_y$ of $z$ on which $s_x{\mathrm{\mid }}_{W_z}=s_y{\mathrm{\mid }}_{W_z}$. The sets $W_z$ cover $V_x\cap V_y$, so [L2] gives $s_x{\mathrm{\mid }}_{V_x\cap V_y}=s_y{\mathrm{\mid }}_{V_x\cap V_y}\mathrm{.}$ Therefore the family $(s_x)$ is compatible on the cover $\{V_x{\}}_{x\in U}$ of $U$.

By [L2], the compatible family of step 2.1 glues to a unique section ${\psi }_U(t)\in \mathcal{F}(U)$ satisfying ${\phi }_U({\psi }_U(t))=t$. Thus ${\phi }_U$ is surjective for every $U$.

Let $V\subseteq U$ and $t\in \mathcal{G}(U)$. Then ${\phi }_V({\psi }_U(t){\mathrm{\mid }}_V)={\phi }_U({\psi }_U(t)){\mathrm{\mid }}_V=t{\mathrm{\mid }}_V={\phi }_V({\psi }_V(t{\mathrm{\mid }}_V))\mathrm{.}$ Step 1.3 gives injectivity of ${\phi }_V$, so ${\psi }_U(t){\mathrm{\mid }}_V={\psi }_V(t{\mathrm{\mid }}_V)$. Hence the maps ${\psi }_U$ commute with restriction and define a morphism of sheaves $\psi :\mathcal{G}\to \mathcal{F}$. By steps 3.1 and 1.3, $\psi$ is a two-sided inverse to $\phi$, so [F1] shows that $\phi$ is an isomorphism. Together with step 1.1 this proves both directions.

For a sheaf $\mathcal{F}$, let $p:E(\mathcal{F})\to X$ be as in [F1]. If $\xi =s_x\in E(\mathcal{F})$, then $\xi \in [s,U]$ for some section $s\in \mathcal{F}(U)$. By construction $p:[s,U]\to U$ is bijective. Its inverse $x^{\mathrm{\prime }}\mapsto s_{x^{\mathrm{\prime }}}$ is continuous because for any smaller basic open $[t,V]\subseteq [s,U]$, the preimage is the open set $\{x^{\mathrm{\prime }}\in U\cap V:s_{x^{\mathrm{\prime }}}=t_{x^{\mathrm{\prime }}}\}$, which is open by the sheaf locality encoded in [L1]. Hence $p{\mathrm{\mid }}_{[s,U]}$ is a homeomorphism onto the open set $U$. Since every point of $E(\mathcal{F})$ lies in some $[s,U]$, $p$ is a local homeomorphism.

Let $q:E\to X$ be a local homeomorphism, and let ${\mathrm{Sec}}_q(U)$ denote its continuous sections over an open set $U$. Restriction of a section is again a section, so ${\mathrm{Sec}}_q$ is a presheaf. If two sections of ${\mathrm{Sec}}_q(U)$ agree on an open cover, they are equal pointwise, so locality holds. If ${\sigma }_i\in {\mathrm{Sec}}_q(U_i)$ are compatible on an open cover $U={\bigcup }_i{U}_i$, define $\sigma (x)={\sigma }_i(x)$ for $x\in U_i$. Compatibility makes this well defined, and continuity is local on the cover because each ${\sigma }_i$ is continuous. Thus [L1] holds, so ${\mathrm{Sec}}_q$ is a sheaf.

A sheaf morphism $\phi :\mathcal{F}\to \mathcal{G}$ induces a map over $X$ $E(\phi ):E(\mathcal{F})⟶E(\mathcal{G}),s_x⟼({\phi }_U(s){)}_x\mathrm{.}$ This is well defined by restriction compatibility. It is continuous: if $E(\phi )(s_x)$ lies in a basic open $[t,V]$, equality of the two germs lets us shrink to an open $W\subseteq U\cap V$ on which ${\phi }_U(s){\mathrm{\mid }}_W=t{\mathrm{\mid }}_W$; then $[s{\mathrm{\mid }}_W,W]$ is a neighbourhood of $s_x$ mapped into $[t,V]$. Identities and compositions are preserved. Conversely, a map $f:E\to E^{\mathrm{\prime }}$ over $X$ between local homeomorphisms sends a section $\sigma$ to $f\circ \sigma$, naturally in the open set, and hence induces a sheaf morphism $\mathrm{Sec}(f)$.

For a sheaf $\mathcal{F}$ and an open set $U$, send $s\in \mathcal{F}(U)$ to the section ${\theta }_U(s):U⟶E(\mathcal{F}),{\theta }_U(s)(x)=s_x\mathrm{.}$ By step 1.1 this is continuous. If ${\theta }_U(s)={\theta }_U(t)$, then $s_x=t_x$ for all $x\in U$, so [L1] gives $s=t$. Conversely, let $\sigma :U\to E(\mathcal{F})$ be a continuous section. For each $x\in U$, choose a basic open $[s_x,V_x]$ containing $\sigma (x)$. Continuity makes $W_x:=U\cap V_x\cap {\sigma }^{-1}([s_x,V_x])$ an open neighbourhood of $x$. Since $p$ is injective on $[s_x,V_x]$ and $p\circ \sigma ={\mathrm{id}}_U$, the restriction $\sigma {\mathrm{\mid }}_{W_x}$ equals $y\mapsto (s_x{)}_y$. These local sections agree on overlaps because they induce the same map $\sigma$. By [L1], they glue to a unique $s\in \mathcal{F}(U)$ with ${\theta }_U(s)=\sigma$. Thus ${\theta }_U$ is a bijection, natural in $U$.

For a local homeomorphism $q:E\to X$, define $\eta :E⟶E({\mathrm{Sec}}_q)$ by sending $e\in E$ to the germ at $x=q(e)$ of any local section through $e$. Such a local section exists because $q$ is a local homeomorphism, and two choices have the same germ because on a smaller common chart they are both the inverse of $q$. On a chart $W\stackrel{\sim }{\to }q(W)$, the map $\eta$ is a homeomorphism from $W$ onto the basic open defined by the inverse section. Hence $\eta$ is an isomorphism over $X$. The formula $E(\phi )(s_x)=({\phi }_U(s){)}_x$ makes the bijections $\theta$ of step 2.1 natural in $\mathcal{F}$, while the formula $\eta (f(e))=(f\circ \sigma {)}_x$ makes $\eta$ natural in $E$. Thus the two functors constructed in steps 1.2 and 1.3 are quasi-inverse equivalences.

Let $\sigma ,\tau \in {\mathcal{F}}^{+}(U)$ and suppose there is an open cover $U={\bigcup }_{\alpha \in A}{W}_{\alpha }$ such that $\sigma {\mathrm{\mid }}_{W_{\alpha }}=\tau {\mathrm{\mid }}_{W_{\alpha }}$ for every $\alpha$. Fix $x\in U$, and choose $\alpha$ with $x\in W_{\alpha }$. Equality on $W_{\alpha }$ means that the two restricted presentations determine the same germ at $x$, so $\sigma$ and $\tau$ have the same germ at $x$. Since this holds for every $x\in U$, the two global presentations are equivalent by [F1]. Therefore $\sigma =\tau$, and ${\mathcal{F}}^{+}$ is separated by [F2].