Comparing Grothendeick topologies

In this posting, we follow Vistoli's notes to discuss how to compare various Grothendieck topologies. Our primary examples will be various small sites of a scheme, and this will let us understand sheaves on one site by understanding them as sheaves on another site.

Sieves. Let $U$ be an object in a category $\mathcal{C}$ and consider a set $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of maps into $U.$ Given any object $T$ in $\mathcal{C},$ define $h_{\mathscr{U}}(T)$ be the set of maps $T \rightarrow U$ such that there is a factorization $T \rightarrow U_{i} \rightarrow U$ by a member of $\mathscr{U}.$ We have $h_{\mathscr{U}}(T) \subset h_{U}(T) = \mathrm{Hom}_{\mathcal{C}}(T, U),$ and this inclusion is functorial in $T.$ That is, we have a subfunctor $h_{\mathscr{U}} \hookrightarrow h_{U},$ and we call $h_{\mathscr{U}}$ the sieve associated with the collection $\mathscr{U}.$

In general, we call any subfunctor of $h_{U}$ a sieve on $U,$ and here is the reason: given a subfunctor $S \hookrightarrow h_{U},$ consider the set $\mathscr{U}_{S} = \bigcup_{T \in \mathrm{Ob}(\mathcal{C})}S(T).$ Then given any object $T$ in $\mathcal{C},$ any map $T \rightarrow U$ that factors through some map $T' \rightarrow U$ in $\mathscr{U}_{S}$ belongs to $S(T).$ That is, we have $h_{\mathscr{U}_{S}}(T) = S(T),$ so $h_{\mathscr{U}_{S}} = S.$ That is, the subfunctor $S$ is the sieve associated with the collection $\mathscr{U}_{S}.$

Sheaf condition via sieves. Now, let $\mathcal{C}$ be equipped with a Grothendieck topology. Given a presheaf $F$ and a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of an object $U,$ denote by $F(\mathscr{U})$ the set of elements $(s_{i})_{i \in I} \in \prod_{i \in I}F(U_{i})$ such that $$s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}}$$ for all $i, j \in I.$ We have a map $F(U) \rightarrow F(\mathscr{U})$ given by $s \mapsto (s|_{U_{i}})_{i \in I}.$ We can immediately note that

• $F$ is a separated presheaf precisely when $F(U) \rightarrow F(\mathscr{U})$ is injective for every cover $\mathscr{U}$ of $U$ and for every object $U$ in $\mathcal{C},$ and
• $F$ is a sheaf precisely when $F(U) \rightarrow F(\mathscr{U})$ is bijective for every cover $\mathscr{U}$ of $U$ and for every object $U$ in $\mathcal{C}.$

Recall the Yoneda map $$\mathrm{Hom}(h_{U}, F) \simeq F(U)$$ given by $\phi \mapsto \phi_{U}(\mathrm{id}_{U}).$ Given a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of $U,$ we consider the map $$\mathrm{Hom}(h_{\mathscr{U}}, F) \rightarrow F(\mathscr{U})$$ given by $\phi \mapsto (\phi_{U_{i}}(U_{i} \rightarrow U))_{i \in I}.$ Since $\phi$ is a natural transformation, we have $$\phi_{U_{i}}(U_{i} \rightarrow U)|_{U_{ij}} = \phi_{U_{ij}}(U_{ij} \rightarrow U) = \phi_{U_{j}}(U_{j} \rightarrow U)|_{U_{ij}},$$ where we wrote $U_{ij} := U_{i} \times_{U} U_{j}.$ Hence, we indeed have $$(\phi_{U_{i}}(U_{i} \rightarrow U))_{i \in I} \in F(\mathscr{U}).$$ We claim that this map is a bijection.

To show injectivity, fix $\phi, \psi \in \mathrm{Hom}(h_{\mathscr{U}}, F)$ such that $$\phi_{U_{i}}(U_{i} \rightarrow U) = \psi_{U_{i}}(U_{i} \rightarrow U)$$ for all $i \in I.$ Given any object $T$ in $\mathcal{C},$ we know any map $T \rightarrow U$ in $h_{\mathscr{U}}(T)$ is factored as $T \rightarrow U_{i} \rightarrow U$ for some $i \in I,$ which let us see that $\phi_{T}(T \rightarrow U)$ is the image of $\phi_{U_{i}}(U_{i} \rightarrow U)$ under $F(U_{i}) \rightarrow F(T).$ Similarly, we note that $\psi_{T}(T \rightarrow U)$ is the image of $\psi_{U_{i}}(U_{i} \rightarrow U)$ under $F(U_{i}) \rightarrow F(T).$ Thus, we must have $$\phi_{T}(T \rightarrow U) = \psi_{T}(T \rightarrow U),$$ showing $\phi = \psi.$ This establishes injectivity.

To show surjectivity, fix any $(s_{i})_{i \in I} \in F(\mathscr{U}).$ Let $T$ be any object in $\mathcal{C},$ and we want to define a map $h_{\mathscr{U}}(T) \rightarrow F(T).$ Fix any $\eta : T \rightarrow U$ in $h_{\mathscr{U}}(T).$ By definition of $h_{\mathscr{U}}(T),$ we have some factorization $$\eta : T \xrightarrow{\eta_{i}} U_{i} \rightarrow U$$ for some $i \in I.$ Then consider $(F\eta_{i})(s_{i}) \in F(T),$ where $F\eta_{i} : F(U_{i}) \rightarrow F(T)$ is induced by $\eta_{i} : T \rightarrow U_{i}.$

Claim. The element $(F\eta_{i})(s_{i}) \in F(T)$ is independent to the choice of the factorization of $T \rightarrow U.$

Proof of Claim. Consider any other factorization $$\eta : T \xrightarrow{\eta_{j}} U_{j} \rightarrow U$$ for some $j \in I.$ Consider the map $(\eta_{i}, \eta_{j}) : T \rightarrow U_{i} \times_{U} U_{j}$ induced by the fiber product. Denoting by $p_{i} : U_{i} \times_{U} U_{j} \rightarrow U_{i}$ and $p_{j} : U_{j} \times_{U} U_{j} \rightarrow U_{j}$ projections, we have $\eta_{i} = p_{i} \circ (\eta_{i}, \eta_{j})$ and $\eta_{j} = p_{j} \circ (\eta_{i}, \eta_{j})$ so that

• $F\eta_{i} = F(\eta_{i}, \eta_{j}) \circ Fp_{i}$ and
• $F\eta_{j} = F(\eta_{i}, \eta_{j}) \circ Fp_{j}$

Since $$(Fp_{i})(s_{i}) = s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}} = (Fp_{j})(s_{j}),$$ this finishes the proof. $\Box$

Hence, given $(s_{i})_{i \in I} \in F(\mathscr{U}),$ we get a well-defined map $$h_{\mathscr{U}}(T) \rightarrow F(T)$$ given by $\eta \mapsto (F\eta_{i})(s_{i}),$ where $\eta_{i} : T \rightarrow U_{i}$ is any map such that $$T \xrightarrow{\eta_{i}} U_{i} \rightarrow U$$ is a factorization of $\eta,$ which always exists by definition. If $\phi : S \rightarrow T$ is any map in $\mathcal{C},$ then for $\eta \in h_{\mathscr{U}}(T),$ we get a factorization $$\eta : T \xrightarrow{\eta_{i}} U_{i} \rightarrow U$$ so that we get the factorization $\eta \circ \phi : S \rightarrow T \rightarrow U_{i} \rightarrow U$ so that $\eta \circ \phi \in h_{\mathscr{U}}(S).$ Now we know that $h_{\mathscr{U}}(S) \rightarrow F(S)$ gives $$\eta \circ \phi \mapsto F(\eta_{i} \circ \phi)(s_{i}) = ((F\phi) \circ (F\eta_{i}))(s_{i}) = (F\phi)((F\eta_{i})(s_{i})),$$ so we have constructed $\psi : h_{\mathscr{U}} \rightarrow F$ as a map of functors. Moreover, we have $\psi_{U_{i}} : h_{\mathscr{U}}(U_{i}) \rightarrow F(U_{i})$ such that $$[U_{i} \rightarrow U] \mapsto (F(\mathrm{id}_{U_{i}})(s_{i}))_{i \in I} = (s_{i})_{i \in I},$$ so this proves the surjectivity of the map $$\mathrm{Hom}(h_{\mathscr{U}}, F) \rightarrow F(\mathscr{U})$$ given by $\psi \mapsto (\psi_{U_{i}}(U_{i} \rightarrow U))_{i \in I},$ so it is a bijection. Note that this bijection restricts to the Yoneda map $\mathrm{Hom}(h_{U}, F) \rightarrow F(U)$ because $$\phi_{U}(\mathrm{id}_{U})|_{U_{i}} = \phi_{U_{i}}(U_{i} \rightarrow U)$$ for all $i \in I.$ Therefore, it follows that:

Corollary. Let $\mathcal{C}$ be a site. Any presheaf $F : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ is a separated presheaf if and only if for any cover $\mathscr{U}$ of any object $U$ in $\mathcal{C},$ the map $$\mathrm{Hom}(h_{U}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{U}}, F)$$ given by $$[h_{U} \rightarrow F] \mapsto [h_{\mathscr{U}} \hookrightarrow h_{U} \rightarrow F]$$ is injective. Moreover, the presheaf $F$ is a sheaf if and only if for any cover $\mathscr{U}$ of any object $U$ in $\mathcal{C},$ the map $\mathrm{Hom}(h_{U}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{U}}, F)$ is bijective.

Refinement. Let $U$ be an object in a category $\mathcal{C}.$ Given a set $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of maps whose target is $U,$ a refinement of $\mathscr{U}$ is a set $\mathscr{U}' = \{U'_{a} \rightarrow U\}_{a \in A}$ of maps with the same target such that every $U'_{a} \rightarrow U$ factors through some $U_{i} \rightarrow U.$ (That is, there is some map $U'_{a} \rightarrow U_{i}$ such that the composition $U'_{a} \rightarrow U_{i} \rightarrow U$ is equal to the map $U'_{a} \rightarrow U.$)

Proposition. Given any two covers $\mathscr{U}$ and $\mathscr{V}$ of an object $U$ in a category $\mathcal{C},$ the cover $\mathscr{V}$ is a refinement of $\mathscr{U}$ if and only if $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}}.$

Proof. Write $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ and $\mathscr{V} = \{V_{j} \rightarrow U\}_{j \in J}.$ If $\mathscr{V}$ is a refinement of $\mathscr{U},$ then $h_{\mathscr{V}}(T) \subset h_{\mathscr{U}}(T)$ for every object $T$ by definition. Conversely, if $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}},$ then for each $j \in J,$ we have $V_{j} \rightarrow U$ in $h_{\mathscr{V}}(V_{j}) \subset h_{\mathscr{U}}(V_{j}),$ so there must be some $i \in I$ such that this map has a factorization $V_{j} \rightarrow U_{i} \rightarrow U.$ $\Box$

The above proposition is easy to prove, but thanks to this, it is easy to notice that refinement gives a partial order on the collection of sets of maps into $U.$ In particular, if $\mathcal{C}$ is a site, then refinements give a partial order on the collection of all the covers of $U.$

Comparing Grothendeick topologies. Let $\mathcal{C}$ be a category and $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ Grothendieck topologies on it. We write $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ and read $\mathcal{T}_{1}$ is subordinate to $\mathcal{T}_{2}$ if every cover of $\mathcal{T}_{1}$ has a refinement in $\mathcal{T}_{2}.$ Note that $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ if and only if for every $\mathscr{U}$ in $\mathcal{T}_{1},$ there is $\mathscr{V}$ in $\mathcal{T}_{2}$ such that $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}}.$

Hence, if $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ and $\mathcal{T}_{2} \prec \mathcal{T}_{3},$ then $\mathcal{T}_{1} \prec \mathcal{T}_{3}.$ Namely, the subordination of Grothendieck topologies is transitive. Thus, it makes sense for us to say that $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ are equivalent if $\mathcal{T}_{1} \prec \mathcal{T}_{2}$ and $\mathcal{T}_{2} \prec \mathcal{T}_{1}.$ This gives an equivalence relation on the Grothendieck topologies on a fixed category $\mathcal{C}.$

Remark. Even though Vistoli does not use this terminology, when $\mathcal{T}_{1} \prec \mathcal{T}_{2},$ I will say $\mathcal{T}_{1}$ is coarser than $\mathcal{T}_{2},$ and $\mathcal{T}_{2}$ is finer than $\mathcal{T}_{2},$ to mean this.

Example. Given any scheme $S,$ on the category $\textbf{Sch}_{S}$ of schemes over $S,$ étale topology is certainly coarser than smooth topology, which has more covers. Vistoli Example 2.51 uses a fact from EGA IV to show that in fact these two Grothendieck topologies are equivalent. I have not checked this proof yet.

Why do we care? Why do we care about comparing Grothendieck topologies? It is because comparison of Grothendieck topologies let us compare the sheaves:

Theorem. Let $\mathcal{C}$ be a category with Grothendieck topologies such that $\mathcal{T}_{1} \prec \mathcal{T}_{2}.$ Then every sheaf on $(\mathcal{C}, \mathcal{T}_{2})$ is a sheaf on $(\mathcal{C}, \mathcal{T}_{1}).$ That is, given two (Grothendieck) topologies, any sheaf on the finer topology is sheaf on the coarser topology.

Proof. Given the hypothesis, let $F$ be a sheaf on $(\mathcal{C}, \mathcal{T}_{2}).$ Given any object $U$ in $\mathcal{C},$ fix any cover $\mathscr{U}$ of $U$ in $(\mathcal{C}, \mathcal{T}_{1}).$ By the hypothesis, there is a cover $\mathscr{V}$ of $U$ in $(\mathcal{C}, \mathcal{T}_{2})$ such that $h_{\mathscr{V}} \hookrightarrow h_{\mathscr{U}}.$ This gives us $$\mathrm{Hom}(h_{U}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{U}}, F) \rightarrow \mathrm{Hom}(h_{\mathscr{V}}, F)$$ given by $$\begin{align*}[h_{U} \rightarrow F] \mapsto [h_{\mathscr{U}} \hookrightarrow h_{U} \rightarrow F]  &\mapsto [h_{\mathscr{U}} \hookrightarrow h_{\mathscr{V}} \hookrightarrow h_{U} \rightarrow F] \\ &= [h_{\mathscr{V}} \hookrightarrow h_{U} \rightarrow F].\end{align*}$$ Since $F$ is a sheaf on $(\mathcal{C}, \mathcal{T}_{2}),$ the above composition is bijective, so the first map must be bijective as well, showing that $F$ is a sheaf on $(\mathcal{C}, \mathcal{T}_{1}).$ $\Box$

General sheafification

In this posting, we follow Vistoli's notes to study a sheafification of a presheaf $F$ on a site $\mathcal{C}$ (i.e., a functor $F : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$). We would like to construct a sheaf $F^{\mathrm{a}} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ equipped with a map $a : F \rightarrow F^{\mathrm{a}}$ of functors such that for any sheaf $G$ and a map $\phi : F^{\mathrm{a}} \rightarrow G,$ there exists a unique map $\phi : F^{\mathrm{a}} \rightarrow G$ with a factorization $F \xrightarrow{a} F^{\mathrm{a}} \xrightarrow{\phi} G$ of $\phi$. It is immediate that a sheafification is unique up to an isomorphism if it ever exists.

Necessary conditions. Instead, we look for a sheaf $F^{\mathrm{a}}$ together with a map $a : F \rightarrow F^{\mathrm{a}}$ such that given any object $U$ of $\mathcal{C},$

1. if $\xi, \eta \in F(U)$ with $a_{U}(\xi) = a_{U}(\eta),$ then there exists a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $\xi|_{U_{i}} = \eta|_{U_{i}}$ for all $i \in I,$ and
2. if $\omega \in F^{a}(U),$ then there exists a cover $\{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $\omega|_{U_{i}} = a_{U_{i}}(\xi_{i})$ for all $i \in I.$

Remark. In a colloquial term, the sheafification of a presheaf remembers local data of each section of the presheaf.

Why are these necessary conditions? Given the condition above, let $\psi : F \rightarrow G$ any map where $G$ is a sheaf. Fix any object $U$ of $\mathcal{C}.$ Given $\omega \in F^{a}(U),$ we may choose a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $\omega|_{U_{i}} = a(\xi_{i})$ for all $i \in I.$ We have $$\begin{align*}a_{U_{ij}}(\xi_{i}|_{U_{ij}}) &= a_{U_{i}}(\xi_{i})|_{U_{ij}} \\ &= \omega|_{U_{ij}} \\ &= a_{U_{j}}(\xi_{j})|_{U_{ij}} \\ &= a_{U_{ij}}(\xi_{j}|_{U_{ij}})\end{align*}$$ in $F^{a}(U_{ij}).$ Thus, there is a cover $\{V_{s} \rightarrow U_{ij}\}_{s \in J}$ such that $$\xi_{i}|_{V_{s}} = \xi_{j}|_{V_{s}}$$ for all $s \in J.$ This implies that $$\psi_{U_{i}}(\xi_{i})|_{V_{s}} = \psi_{V_{s}}(\xi_{i}|_{V_{s}}) = \psi_{V_{s}}(\xi_{j}|_{V_{s}}) = \psi_{U_{j}}(\xi_{j})|_{V_{s}}$$ for all $s \in J,$ so we must have $$\psi_{U_{i}}(\xi_{i})|_{U_{ij}} = \psi_{U_{j}}(\xi_{j})|_{U_{ij}}$$ in $G(U_{ij})$ for $i, j \in I.$ Since $G$ is a sheaf, this implies that there is a unique $\tilde{\psi}_{\mathscr{U}}(\omega) \in G(U)$ such that $\tilde{\psi}_{\mathscr{U}}(\omega)|_{U_{i}} = \psi_{U_{i}}(\xi_{i})$ for all $i \in I.$ If we have two covers of $U,$ we may take their refinements, and since $G$ is a sheaf, this implies that $\tilde{\psi}_{\mathscr{U}}(\omega)$ does not depend on the choice of the cover $\mathscr{U},$ so we may write $\tilde{\psi}_{U}(\omega) := \tilde{\psi}_{\mathscr{U}}(\omega)$ instead. Hence, we have defined a map $\tilde{\psi}_{U} : F^{a}(U) \rightarrow G(U).$ If $V \rightarrow U$ is a map in $\mathcal{C}$ and $s \in F^{a}(U),$ then taking a cover $\{U_{i} \rightarrow U\}_{i \in I}$ for $U$ to get a cover $\{V_{i} \rightarrow V\}_{i \in I}$ for $V$ where $V_{i} = V \times_{U} U_{i},$ we have $$(\tilde{\psi}_{U}(s)|_{V})|_{V_{i}} = \tilde{\psi}_{U}(s)|_{V_{i}} = \tilde{\psi}_{V_{i}}(s) = \tilde{\psi}_{V}(s)|_{V_{i}}$$ for all $i \in I,$ so $\tilde{\psi}_{U}(s)|_{V} = \tilde{\psi}_{V}(s).$ Thus, our definition gives a map $\tilde{\psi} : F^{a} \rightarrow G$ of sheaves. The definition of $\tilde{\psi}$ immediately tells us that $\tilde{\psi} \circ a = \psi.$ To show uniqueness, say we have another $\psi' : F^{a} \rightarrow G$ such that $\psi' \circ a = \psi.$ Then given any $\omega \in F^{a}(U),$ we choose a cover $\{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $a_{U_{i}}(\xi_{i}) = \omega|_{U_{i}}$ for $i \in I.$ This implies that $$\begin{align*}\psi'_{U}(\omega)|_{U_{i}} &= \psi'_{U_{i}}(\omega|_{U_{i}}) \\ &= \psi'_{U_{i}}(a_{U_{i}}(\xi_{i})) \\ &= \psi_{U_{i}}(\xi_{i}) \\ &= \tilde{\psi}_{U_{i}}(a_{U_{i}}(\xi_{i})) \\ &= \tilde{\psi}_{U_{i}}(\omega|_{U_{i}}) \\ &= \tilde{\psi}_{U}(\omega)|_{U_{i}}\end{align*}$$ for all $i \in I,$ so $\psi'_{U}(\omega) = \tilde{\psi}_{U}(\omega).$ Hence, we have $\psi' = \tilde{\psi},$ showing the uniqueness.

Sheafifications valued in different categories. The discussion above applies even when we replace $\textbf{Set}$ with $\textbf{Ab}$ or the category of modules over a ring.

Construction. For any $s, t \in F(U),$ we write $s \sim t$ if there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I.$ This is an equivalence relation where we can check transitivity by taking the joint cover of two given covers.

Define $F^{\mathrm{s}}(U)$ be the set of equivalence classes of elements in $F(U)$ with respect to this equivalence relation. Given any morphism $V \rightarrow U$ in $\mathcal{C},$ if $s \sim t$ in $F(U),$ we have a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I,$ so $(s|_{V})|_{V_{i}} = (t|_{V})|_{V_{i}}$ for all $i \in I,$ where $\{V \times_{U} U_{i} = V_{i} \rightarrow V\}_{i \in I}$ is a cover for $V.$ Hence, we have $s|_{V} \sim t|_{V}$ in $F(V).$ This lets us define $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ by $[s] \mapsto [s|_{V}].$ This immediately makes $F^{\mathrm{s}} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ a functor (i.e., a presheaf). If $\{U_{i} \rightarrow U\}_{i \in I}$ is a cover and $s, t \in F(U)$ with $s|_{U_{i}} \sim t|_{U_{i}} \in F(U_{i})$ for all $i \in I,$ then we may take a cover $\{V_{ij} \rightarrow U_{i}\}_{j \in J_{i}}$ for each $i \in I$ such that $s|_{V_{ij}} = t|_{V_{ij}}$ for all $j \in J_{i}.$ Since $\{V_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}}$ is a cover for $U,$ we have $s \sim t$ in $F(U).$ This implies that $F^{\mathrm{s}}$ is a separated presheaf.

Given an object $U$ of $\mathcal{C},$ we consider the set of pairs $(\mathscr{U}, \boldsymbol{s}),$ where $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ is a cover of $U$ and $\boldsymbol{s} = ([s_{i}])_{i \in I} \in \prod_{i \in I}F^{\mathrm{s}}(U_{i})$ such that $[s_{i}|_{U_{ij}}] = [s_{j}|_{U_{ij}}] \in F^{\mathrm{s}}(U_{ij})$ for all $i, j \in I$ where $U_{ij} = U_{i} \times_{U} U_{j}.$ For two pairs $(\mathscr{U}, \boldsymbol{s})$ and $(\mathscr{V}, \boldsymbol{t}),$ we write $$(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{V}, \boldsymbol{t})$$ if $$[s_{i}|_{U_{i} \times_{U} V_{j}}] = [t_{j}|_{U_{i} \times_{U} V_{j}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} V_{j})$ for all $i \in I$ and $j \in J,$ where we wrote 

• $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I},$
• $\mathscr{V} = \{V_{j} \rightarrow U\}_{j \in J},$
• $\boldsymbol{s} = ([s_{i}])_{i \in I},$
• $\boldsymbol{t} = ([t_{j}])_{j \in J}.$

We claim that this gives an equivalence relation among such pairs. Reflexivity is evident, and for symmetry, we just need to use the isomorphisms $V_{j} \times_{U} U_{i} \simeq U_{i} \times_{U} V_{j}$ because the covers always include isomorphisms and we can compose covers to get another. For transitivity, suppose that $(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{V}, \boldsymbol{t})$ and $(\mathscr{V}, \boldsymbol{t}) \sim (\mathscr{W}, \boldsymbol{u}),$ where we write  

• $\mathscr{W} = \{W_{k} \rightarrow U\}_{k \in K},$
• $\boldsymbol{u} = ([u_{k}])_{k \in K}.$

We have $$[s_{i}|_{U_{i} \times_{U} V_{j}}] = [t_{j}|_{U_{i} \times_{U} V_{j}}]$$ for all $i \in I$ and $j \in J,$ while $$[t_{j}|_{V_{j} \times_{U} W_{k}}] = [u_{k}|_{V_{j} \times_{U} W_{k}}]$$ for all $j \in J$ and $k \in K.$ This gives us $$[s_{i}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [t_{j}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [u_{k}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}].$$ Running over $j \in V,$ since $F^{\mathrm{s}}$ is a separated presheaf, we necessarily have $$[s_{i}|_{U_{i} \times_{U} W_{k}}] = [u_{k}|_{U_{i} \times_{U} W_{k}}].$$ This shows the transitivity, so we indeed get an equivalence relation.

We define $F^{\mathrm{a}}(U)$ to be the set of equivalence classes $[(\mathscr{U}, \boldsymbol{s})]$ of the pairs we have discussed above. Given a morphism $V \rightarrow U$ in $\mathcal{C},$ we define $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ by $[(\mathscr{U}, \boldsymbol{s})] \mapsto [(V \times_{U} \mathscr{U}, \boldsymbol{s}^{*})],$ where $V \times_{U} \mathscr{U} = \{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ and $\boldsymbol{s}^{*} = (s_{i}|_{V \times U_{i}})_{i \in I}.$ Given $(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{W}, \boldsymbol{t}),$ we have $$[s_{i}|_{U_{i} \times_{U} W_{j}}] = [t_{j}|_{U_{i} \times_{U} W_{j}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} W_{j}),$ for all $i \in I$ and $j \in J.$ Since $$(V \times_{U} U_{i}) \times_{V} (V \times_{U} W_{j}) \simeq V \times_{U} U_{i} \times_{U} W_{j},$$ having $$[s_{i}|_{V \times_{U} U_{i} \times_{U} W_{j}}] = [t_{j}|_{V \times_{U} U_{i} \times_{U} W_{j}}]$$ is enough to check that the restrictions $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ are well-defined.

Given a cover $\{U_{i} \rightarrow U\}_{i \in I},$ fix any $[(\mathscr{V}, \boldsymbol{s})], [(\mathscr{W}, \boldsymbol{t})] \in F^{a}(U)$ such that $$[(U_{i} \times_{U} \mathscr{V}, \boldsymbol{s}^{*})] = [(U_{i} \times_{U}\mathscr{W}, \boldsymbol{t}^{*})]$$ for all $i \in I.$ Since $$(U_{i} \times_{U} V_{j}) \times_{U_{i}} (U_{i} \times_{U} W_{k}) \simeq U_{i} \times_{U} V_{j} \times_{U} W_{k},$$ we necessarily have $$[s_{j}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [t_{k}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} V_{j} \times_{U} W_{k})$. Ranging over all $i \in I,$ using the fact that $F^{\mathrm{s}}$ is a separated presheaf, we have $$[s_{j}|_{V_{j} \times_{U} W_{k}}] = [t_{k}|_{V_{j} \times_{U} W_{k}}]$$ in $F^{\mathrm{s}}(V_{j} \times_{U} W_{k})$. This implies that $$[(\mathscr{V}, \boldsymbol{s})] = [(\mathscr{W}, \boldsymbol{t})].$$ This shows that $F^{\mathrm{a}}$ is a separated presheaf.

To show that $F^{\mathrm{a}}$ is a sheaf, fix $[(\mathscr{V}_{i}, \boldsymbol{s}_{i})] \in F^{\mathrm{a}}(U_{i})$ for $i \in I$ such that $$[(U_{i} \times_{U} U_{j} \times_{U} \mathscr{V}_{i} , \boldsymbol{s}_{i}^{*})] = [(U_{i} \times_{U} U_{j} \times_{U} \mathscr{V}_{j}, \boldsymbol{s}_{j}^{*})].$$ We have $$(U_{i} \times_{U} U_{j} \times_{U} V_{i,l}) \times_{U_{i} \times_{U} U_{j}} (U_{i} \times_{U} U_{j} \times_{U} V_{j,m}) \simeq U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm},$$ so the above identity implies that we have $$s_{i,l} |_{U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm}} = s_{j,m} |_{U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm}},$$ but then this makes $[(\mathscr{V}, \boldsymbol{s})] \in F^{\mathrm{a}}(U)$ given by $\mathscr{V} = \{V_{il} \rightarrow U_{i} \rightarrow U\}_{i,l}$ and $\boldsymbol{s} = ([s_{i,l}])_{i,l} \in \prod_{i,l} F^{\mathrm{s}}(V_{il}).$ This shows that $F^{\mathrm{a}}$ is a sheaf.

We define $a_{U} : F(U) \rightarrow F^{\mathrm{a}}(U)$ by $s \mapsto [(\{\mathrm{id}_{U}\}, (s))].$ It is immediate that this gives rise to a map $a : F \rightarrow F^{\mathrm{a}}$ of functors. Given $s, t \in F(U)$ with $$[(\{\mathrm{id}_{U}\}, (s))] = [(\{\mathrm{id}_{U}\}, (t))],$$ we have $[s] = [t] \in F^{\mathrm{s}}(U),$ so by our definition, there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I.$ Any given $[(\mathscr{U}, \textbf{s})] \in F^{\mathrm{a}}(U)$ comes with the cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ and $\boldsymbol{s} = ([s_{i}])_{i \in I} \in \prod_{i \in I}F^{\mathrm{s}}(U_{i})$ so that $$a_{U_{i}}(s_{i}) = [(\{\mathrm{id}_{U_{i}}\}, (s_{i}))] = [(U_{i} \times_{U} \mathscr{U}, \boldsymbol{s}^{*})]| = [(\mathscr{U}, \boldsymbol{s})]|_{U_{i}},$$ where we get the second identity because $$U_{i} \times_{U_{i}} (U_{i} \times_{U} U_{j}) \simeq U_{i} \times_{U} U_{j}$$ and $s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}}$ for all $j \in J.$ Thus, we are done checking the necessary conditions for $a : F \rightarrow F^{\mathrm{a}}$ to be a sheafification of $F.$