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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