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
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
- 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
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