Lemma 7.10.1.slogan Let \mathcal{F} : \mathcal{I} \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) be a diagram. Then \mathop{\mathrm{lim}}\nolimits _\mathcal {I} \mathcal{F} exists and is equal to the limit in the category of presheaves.
7.10 Sheafification
In order to define the sheafification we study the zeroth Čech cohomology group of a covering and its functoriality properties.
Let \mathcal{F} be a presheaf of sets on \mathcal{C}, and let \mathcal{U} = \{ U_ i \to U\} _{i \in I} be a covering of \mathcal{C}. Let us use the notation \mathcal{F}(\mathcal{U}) to indicate the equalizer
As we will see later, this is the zeroth Čech cohomology of \mathcal{F} over U with respect to the covering \mathcal{U}. A small remark is that we can define H^0(\mathcal{U}, \mathcal{F}) as soon as all the morphisms U_ i \to U are representable, i.e., \mathcal{U} need not be a covering of the site. There is a canonical map \mathcal{F}(U) \to H^0(\mathcal{U}, \mathcal{F}). It is clear that a morphism of coverings \mathcal{U} \to \mathcal{V} induces commutative diagrams
This in turn produces a map H^0(\mathcal{V}, \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F}), compatible with the map \mathcal{F}(V) \to \mathcal{F}(U).
By construction, a presheaf \mathcal{F} is a sheaf if and only if for every covering \mathcal{U} of \mathcal{C} the natural map \mathcal{F}(U) \to H^0(\mathcal{U}, \mathcal{F}) is bijective. We will use this notion to prove the following simple lemma about limits of sheaves.
Proof. Let \mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i be the limit as a presheaf. We will show that this is a sheaf and then it will trivially follow that it is a limit in the category of sheaves. To prove the sheaf property, let \mathcal{V} = \{ V_ j \to V\} _{j\in J} be a covering. Let (s_ j)_{j\in J} be an element of H^0(\mathcal{V}, \mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i). Using the projection maps we get elements (s_{j, i})_{j\in J} in H^0(\mathcal{V}, \mathcal{F}_ i). By the sheaf property for \mathcal{F}_ i we see that there is a unique s_ i \in \mathcal{F}_ i(V) such that s_{j, i} = s_ i|_{V_ j}. Let \phi : i \to i' be a morphism of the index category. We would like to show that \mathcal{F}(\phi ) : \mathcal{F}_ i \to \mathcal{F}_{i'} maps s_ i to s_{i'}. We know this is true for the sections s_{i, j} and s_{i', j} for all j and hence by the sheaf property for \mathcal{F}_{i'} this is true. At this point we have an element s = (s_ i)_{i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})} of (\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i)(V). We leave it to the reader to see this element has the required property that s_ j = s|_{V_ j}. \square
Example 7.10.2. A particular example is the limit over the empty diagram. This gives the final object in the category of (pre)sheaves. It is the presheaf that associates to each object U of \mathcal{C} a singleton set, with unique restriction mappings and moreover this presheaf is a sheaf. We often denote this sheaf by *.
Let \mathcal{J}_ U be the category of all coverings of U. In other words, the objects of \mathcal{J}_ U are the coverings of U in \mathcal{C}, and the morphisms are the refinements. By our conventions on sites this is indeed a category, i.e., the collection of objects and morphisms forms a set. Note that \mathop{\mathrm{Ob}}\nolimits (\mathcal{J}_ U) is not empty since \{ \text{id}_ U\} is an object of it. According to the remarks above the construction \mathcal{U} \mapsto H^0(\mathcal{U}, \mathcal{F}) is a contravariant functor on \mathcal{J}_ U. We define
See Categories, Section 4.14 for a discussion of limits and colimits. We point out that later we will see that \mathcal{F}^{+}(U) is the zeroth Čech cohomology of \mathcal{F} over U.
Before we say more about the structure of the colimit, we turn the collection of sets \mathcal{F}^{+}(U), U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) into a presheaf. Namely, let V \to U be a morphism of \mathcal{C}. By the axioms of a site there is a functor1
Note that the projection maps furnish a functorial morphism of coverings \{ U_ i \times _ U V \to V\} \to \{ U_ i \to U\} and hence, by the construction above, a functorial map of sets H^0(\{ U_ i \to U\} , \mathcal{F}) \to H^0(\{ U_ i \times _ U V \to V\} , \mathcal{F}). In other words, there is a transformation of functors from H^0(-, \mathcal{F}) : \mathcal{J}_ U^{opp} \to \textit{Sets} to the composition \mathcal{J}_ U^{opp} \to \mathcal{J}_ V^{opp} \xrightarrow {H^0(-, \mathcal{F})} \textit{Sets}. Hence by generalities of colimits we obtain a canonical map \mathcal{F}^+(U) \to \mathcal{F}^+(V). In terms of the description of the set \mathcal{F}^+(U) above, it just takes the element associated with s = (s_ i) \in H^0(\{ U_ i \to U\} , \mathcal{F}) to the element associated with (s_ i|_{V \times _ U U_ i}) \in H^0(\{ U_ i \times _ U V \to V\} , \mathcal{F}).
Lemma 7.10.3. The constructions above define a presheaf \mathcal{F}^+ together with a canonical map of presheaves \mathcal{F} \to \mathcal{F}^+.
Proof. All we have to do is to show that given morphisms W \to V \to U the composition \mathcal{F}^+(U) \to \mathcal{F}^+(V) \to \mathcal{F}^+(W) equals the map \mathcal{F}^+(U) \to \mathcal{F}^+(W). This can be shown directly by verifying that, given a covering \{ U_ i \to U\} and s = (s_ i) \in H^0(\{ U_ i \to U\} , \mathcal{F}), we have canonically W \times _ U U_ i \cong W \times _ V (V \times _ U U_ i), and s_ i|_{W \times _ U U_ i} corresponds to (s_ i|_{V \times _ U U_ i})|_{W \times _ V (V \times _ U U_ i)} via this isomorphism. \square
More indirectly, the result of Lemma 7.10.6 shows that we may pullback an element s as above via any morphism from any covering of W to \{ U_ i \to U\} and we will always end up with the same element in \mathcal{F}^+(W).
Lemma 7.10.4. The association \mathcal{F} \mapsto (\mathcal{F} \to \mathcal{F}^+) is a functor.
Proof. Instead of proving this we state exactly what needs to be proven. Let \mathcal{F} \to \mathcal{G} be a map of presheaves. Prove the commutativity of:
The next two lemmas imply that the colimits above are colimits over a directed set.
Lemma 7.10.5. Given a pair of coverings \{ U_ i \to U\} and \{ V_ j \to U\} of a given object U of the site \mathcal{C}, there exists a covering which is a common refinement.
Proof. Since \mathcal{C} is a site we have that for every i the family \{ V_ j \times _ U U_ i \to U_ i\} _ j is a covering. And, then another axiom implies that \{ V_ j \times _ U U_ i \to U\} _{i, j} is a covering of U. Clearly this covering refines both given coverings. \square
Lemma 7.10.6. Any two morphisms f, g: \mathcal{U} \to \mathcal{V} of coverings inducing the same morphism U \to V induce the same map H^0(\mathcal{V}, \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F}).
Proof. Let \mathcal{U} = \{ U_ i \to U\} _{i\in I} and \mathcal{V} = \{ V_ j \to V\} _{j\in J}. The morphism f consists of a map U\to V, a map \alpha : I \to J and maps f_ i : U_ i \to V_{\alpha (i)}. Likewise, g determines a map \beta : I \to J and maps g_ i : U_ i \to V_{\beta (i)}. As f and g induce the same map U\to V, the diagram
is commutative for every i\in I. Hence f and g factor through the fibre product
Now let s = (s_ j)_ j \in H^0(\mathcal{V}, \mathcal{F}). Then for all i\in I:
where the middle equality is given by the definition of H^0(\mathcal{V}, \mathcal{F}). This shows that the maps H^0(\mathcal{V}, \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F}) induced by f and g are equal. \square
Remark 7.10.7. In particular this lemma shows that if \mathcal{U} is a refinement of \mathcal{V}, and if \mathcal{V} is a refinement of \mathcal{U}, then there is a canonical identification H^0(\mathcal{U}, \mathcal{F}) = H^0(\mathcal{V}, \mathcal{F}).
From these two lemmas, and the fact that \mathcal{J}_ U is nonempty, it follows that the diagram H^0(-, \mathcal{F}) : \mathcal{J}_ U^{opp} \to \textit{Sets} is filtered, see Categories, Definition 4.19.1. Hence, by Categories, Section 4.19 the colimit \mathcal{F}^{+}(U) may be described in the following straightforward manner. Namely, every element in the set \mathcal{F}^{+}(U) arises from an element s \in H^0(\mathcal{U}, \mathcal{F}) for some covering \mathcal{U} of U. Given a second element s' \in H^0(\mathcal{U}', \mathcal{F}) then s and s' determine the same element of the colimit if and only if there exists a covering \mathcal{V} of U and refinements f : \mathcal{V} \to \mathcal{U} and f' : \mathcal{V} \to \mathcal{U}' such that f^*s = (f')^*s' in H^0(\mathcal{V}, \mathcal{F}). Since the trivial covering \{ \text{id}_ U\} is an object of \mathcal{J}_ U we get a canonical map \mathcal{F}(U) \to \mathcal{F}^+(U).
Lemma 7.10.8. The map \theta : \mathcal{F} \to \mathcal{F}^+ has the following property: For every object U of \mathcal{C} and every section s \in \mathcal{F}^+(U) there exists a covering \{ U_ i \to U\} such that s|_{U_ i} is in the image of \theta : \mathcal{F}(U_ i) \to \mathcal{F}^{+}(U_ i).
Proof. Namely, let \{ U_ i \to U\} be a covering such that s arises from the element (s_ i) \in H^0(\{ U_ i \to U\} , \mathcal{F}). According to Lemma 7.10.6 we may consider the covering \{ U_ i \to U_ i\} and the (obvious) morphism of coverings \{ U_ i \to U_ i\} \to \{ U_ i \to U\} to compute the pullback of s to an element of \mathcal{F}^+(U_ i). And indeed, using this covering we get exactly \theta (s_ i) for the restriction of s to U_ i. \square
Definition 7.10.9. We say that a presheaf of sets \mathcal{F} on a site \mathcal{C} is separated if, for all coverings of \{ U_ i \rightarrow U\} , the map \mathcal{F}(U) \to \prod \mathcal{F}(U_ i) is injective.
Theorem 7.10.10. With \mathcal{F} as above
The presheaf \mathcal{F}^+ is separated.
If \mathcal{F} is separated, then \mathcal{F}^+ is a sheaf and the map of presheaves \mathcal{F} \to \mathcal{F}^+ is injective.
If \mathcal{F} is a sheaf, then \mathcal{F} \to \mathcal{F}^+ is an isomorphism.
The presheaf \mathcal{F}^{++} is always a sheaf.
Proof. Proof of (1). Suppose that s, s' \in \mathcal{F}^+(U) and suppose that there exists some covering \{ U_ i \to U\} such that s|_{U_ i} = s'|_{U_ i} for all i. We now have three coverings of U: the covering \{ U_ i \to U\} above, a covering \mathcal{U} for s as in Lemma 7.10.8, and a similar covering \mathcal{U}' for s'. By Lemma 7.10.5, we can find a common refinement, say \{ W_ j \to U\} . This means we have s_ j, s'_ j \in \mathcal{F}(W_ j) such that s|_{W_ j} = \theta (s_ j), similarly for s'|_{W_ j}, and such that \theta (s_ j) = \theta (s'_ j). This last equality means that there exists some covering \{ W_{jk} \to W_ j\} such that s_ j|_{W_{jk}} = s'_ j|_{W_{jk}}. Then since \{ W_{jk} \to U\} is a covering we see that s, s' map to the same element of H^0(\{ W_{jk} \to U\} , \mathcal{F}) as desired.
Proof of (2). It is clear that \mathcal{F} \to \mathcal{F}^+ is injective because all the maps \mathcal{F}(U) \to H^0(\mathcal{U}, \mathcal{F}) are injective. It is also clear that, if \mathcal{U} \to \mathcal{U}' is a refinement, then H^0(\mathcal{U}', \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F}) is injective. Now, suppose that \{ U_ i \to U\} is a covering, and let (s_ i) be a family of elements of \mathcal{F}^+(U_ i) satisfying the sheaf condition s_ i|_{U_ i \times _ U U_{i'}} = s_{i'}|_{U_ i \times _ U U_{i'}} for all i, i' \in I. Choose coverings (as in Lemma 7.10.8) \{ U_{ij} \to U_ i\} such that s_ i|_{U_{ij}} is the image of the (unique) element s_{ij} \in \mathcal{F}(U_{ij}). The sheaf condition implies that s_{ij} and s_{i'j'} agree over U_{ij} \times _ U U_{i'j'} because it maps to U_ i \times _ U U_{i'} and we have the equality there. Hence (s_{ij}) \in H^0(\{ U_{ij} \to U\} , \mathcal{F}) gives rise to an element s \in \mathcal{F}^+(U). We leave it to the reader to verify that s|_{U_ i} = s_ i.
Proof of (3). This is immediate from the definitions because the sheaf property says exactly that every map \mathcal{F} \to H^0(\mathcal{U}, \mathcal{F}) is bijective (for every covering \mathcal{U} of U).
Statement (4) is now obvious. \square
Definition 7.10.11. Let \mathcal{C} be a site and let \mathcal{F} be a presheaf of sets on \mathcal{C}. The sheaf \mathcal{F}^\# := \mathcal{F}^{++} together with the canonical map \mathcal{F} \to \mathcal{F}^\# is called the sheaf associated to \mathcal{F}.
Proposition 7.10.12. The canonical map \mathcal{F} \to \mathcal{F}^\# has the following universal property: For any map \mathcal{F} \to \mathcal{G}, where \mathcal{G} is a sheaf of sets, there is a unique map \mathcal{F}^\# \to \mathcal{G} such that \mathcal{F} \to \mathcal{F}^\# \to \mathcal{G} equals the given map.
Proof. By Lemma 7.10.4 we get a commutative diagram
and by Theorem 7.10.10 the lower horizontal maps are isomorphisms. The uniqueness follows from Lemma 7.10.8 which says that every section of \mathcal{F}^\# locally comes from sections of \mathcal{F}. \square
It is clear from this result that the functor \mathcal{F} \mapsto (\mathcal{F} \to \mathcal{F}^\# ) is unique up to unique isomorphism of functors. Actually, let us temporarily denote i : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \textit{PSh}(\mathcal{C}) the functor of inclusion. The result above actually says that
In other words, the functor of sheafification is the left adjoint to the inclusion functor i. We finish this section with a couple of lemmas.
Lemma 7.10.13.slogan Let \mathcal{F} : \mathcal{I} \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) be a diagram. Then \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathcal{F} exists and is the sheafification of the colimit in the category of presheaves.
Proof. Since the sheafification functor is a left adjoint it commutes with all colimits, see Categories, Lemma 4.24.5. Hence, since \textit{PSh}(\mathcal{C}) has colimits, we deduce that \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) has colimits (which are the sheafifications of the colimits in presheaves). \square
Lemma 7.10.14. The functor \textit{PSh}(\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{F} \mapsto \mathcal{F}^\# is exact.
Proof. Since it is a left adjoint it is right exact, see Categories, Lemma 4.24.6. On the other hand, by Lemmas 7.10.5 and Lemma 7.10.6 the colimits in the construction of \mathcal{F}^+ are really over the directed set \mathop{\mathrm{Ob}}\nolimits (\mathcal{J}_ U) where \mathcal{U} \geq \mathcal{U}' if and only if \mathcal{U} is a refinement of \mathcal{U}'. Hence by Categories, Lemma 4.19.2 we see that \mathcal{F} \to \mathcal{F}^+ commutes with finite limits (as a functor from presheaves to presheaves). Then we conclude using Lemma 7.10.1. \square
Lemma 7.10.15. Let \mathcal{C} be a site. Let \mathcal{F} be a presheaf of sets on \mathcal{C}. Denote \theta ^2 : \mathcal{F} \to \mathcal{F}^\# the canonical map of \mathcal{F} into its sheafification. Let U be an object of \mathcal{C}. Let s \in \mathcal{F}^\# (U). There exists a covering \{ U_ i \to U\} and sections s_ i \in \mathcal{F}(U_ i) such that
s|_{U_ i} = \theta ^2(s_ i), and
for every i, j there exists a covering \{ U_{ijk} \to U_ i \times _ U U_ j\} of \mathcal{C} such that the pullbacks of s_ i and s_ j to each U_{ijk} agree.
Conversely, given any covering \{ U_ i \to U\} , elements s_ i \in \mathcal{F}(U_ i) such that (2) holds, then there exists a unique section s \in \mathcal{F}^\# (U) such that (1) holds.
Proof. Omitted. \square
Lemma 7.10.16. Let \mathcal{C} be a site. Let \mathcal{F} \to \mathcal{G} be a map of presheaves of sets on \mathcal{C}. Denote \mathcal{B} the set of U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) such that \mathcal{F}(U) \to \mathcal{G}(U) is bijective. If every object of \mathcal{C} has a covering by elements of \mathcal{B}, then \mathcal{F}^\# \to \mathcal{G}^\# is an isomorphism.
Proof. Let U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}). Let us prove that \mathcal{F}^\# (U) \to \mathcal{G}^\# (U) is surjective. To do this we will use Lemma 7.10.15 without further mention. For any s \in \mathcal{G}^\# (U) there exists a covering \{ U_ i \to U\} and sections s_ i \in \mathcal{G}(U_ i) such that
s|_{U_ i} is the image of s_ i via \mathcal{G} \to \mathcal{G}^\# , and
for every i, j there exists a covering \{ U_{ijk} \to U_ i \times _ U U_ j\} of \mathcal{D} such that the pullbacks of s_ i and s_ j to each U_{ijk} agree.
By assumption, for each i we may choose a covering \{ U_{i, a} \to U_ i\} with U_{i, a} \in \mathcal{B}. Then \{ U_{i, a} \to U\} is a covering. Denoting s_{i, a} the image of s_ i in \mathcal{G}(U_{i, a}) we see that the pullbacks of s_{i, a} and s_{j, b} to the members of the covering
agree. Hence we may assume that U_ i \in \mathcal{B}. Repeating the argument, we may also assume U_{ijk} \in \mathcal{B} for all i, j, k (details omitted). Then since \mathcal{F}(U_ i) \to \mathcal{G}(U_ i) is bijective by our definition of \mathcal{B} in the statement of the lemma, we get unique t_ i \in \mathcal{F}(U_ i) mapping to s_ i. The pullbacks of t_ i and t_ j to each U_{ijk} agree in \mathcal{F}(U_{ijk}) because U_{ijk} \in \mathcal{B} and because we have the agreement for t_ i and t_ j. Then the lemma tells us there exists a unique t \in \mathcal{F}^\# (U) such that t|_{U_ i} is the image of t_ i. It follows that t maps to s. We omit the proof that \mathcal{F}^\# (U) \to \mathcal{G}^\# (U) is injective. \square
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