## 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

$H^0(\mathcal{U}, \mathcal{F}) = \{ (s_ i)_{i\in I} \in \prod \nolimits _ i \mathcal{F}(U_ i) \mid s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j} \ \forall i, j \in I \} .$

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

$\xymatrix{ & U_ i \ar[rr] & & V_{\alpha (i)} \\ U_ i \times _ U U_ j \ar[rr] \ar[ur] \ar[dr] & & V_{\alpha (i)} \times _ V V_{\alpha (j)} \ar[ur] \ar[dr] & \\ & U_ j \ar[rr] & & V_{\alpha (j)} }.$

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.

Lemma 7.10.1. 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.

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

$\mathcal{F}^{+}(U) = \mathop{\mathrm{colim}}\nolimits _{\mathcal{J}_ U^{opp}} H^0(\mathcal{U}, \mathcal{F})$

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

$\mathcal{J}_ U \longrightarrow \mathcal{J}_ V, \quad \{ U_ i \to U\} \longmapsto \{ U_ i \times _ U V \to V\} .$

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:

$\xymatrix{ \mathcal{F} \ar[r] \ar[d] & \mathcal{F}^{+} \ar[d] \\ \mathcal{G} \ar[r] & \mathcal{G}^{+} }$
$\square$

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

$\xymatrix{ & V_{\alpha (i)} \ar[dr] \\ U_ i \ar[ur]^{f_ i} \ar[dr]_{g_ i} & & V \\ & V_{\beta (i)} \ar[ur] }$

is commutative for every $i\in I$. Hence $f$ and $g$ factor through the fibre product

$\xymatrix{ & V_{\alpha (i)} \\ U_ i \ar[r]^-\varphi \ar[ur]^{f_ i} \ar[dr]_{g_ i} & V_{\alpha (i)} \times _ V V_{\beta (i)} \ar[u]_{\text{pr}_1} \ar[d]^{\text{pr}_2} \\ & V_{\beta (i)}. }$

Now let $s = (s_ j)_ j \in H^0(\mathcal{V}, \mathcal{F})$. Then for all $i\in I$:

$(f^*s)_ i = f_ i^*(s_{\alpha (i)}) = \varphi ^*\text{pr}_1^*(s_{\alpha (i)}) = \varphi ^*\text{pr}_2^*(s_{\beta (i)}) = g_ i^*(s_{\beta (i)}) = (g^*s)_ i,$

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

1. The presheaf $\mathcal{F}^+$ is separated.

2. If $\mathcal{F}$ is separated, then $\mathcal{F}^+$ is a sheaf and the map of presheaves $\mathcal{F} \to \mathcal{F}^+$ is injective.

3. If $\mathcal{F}$ is a sheaf, then $\mathcal{F} \to \mathcal{F}^+$ is an isomorphism.

4. 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 $s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j}$ for all $i, j \in I$. 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

$\xymatrix{ \mathcal{F} \ar[r] \ar[d] & \mathcal{F}^{+} \ar[r] \ar[d] & \mathcal{F}^{++} \ar[d] \\ \mathcal{G} \ar[r] & \mathcal{G}^{+} \ar[r] & \mathcal{G}^{++} }$

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

$\mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{F}, i(\mathcal{G})) = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}^\# , \mathcal{G}).$

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. 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

1. $s|_{U_ i} = \theta ^2(s_ i)$, and

2. for every $i, j$ there exists a covering $\{ U_{ijk} \to U_ i \times _ U U_ j\}$ of $\mathcal{C}$ such that the pullback 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. $\square$

 This construction actually involves a choice of the fibre products $U_ i \times _ U V$ and hence the axiom of choice. The resulting map does not depend on the choices made, see below.

Comment #2953 by Ko Aoki on

Typos: Some equations says "$H^0(-, \mathcal{F}) : \mathcal{J}_U \to \textit{Sets}$", but the correct source of $H^0(-, \mathcal{F})$ is $\mathcal{J}_U^{opp}$.

Comment #3979 by Lucy on

In example 7.10.2, it should be (U)={} instead of ∅ if U has an empty covering, as explained in https://stacks.math.columbia.edu/tag/04B3

Comment #4332 by ExcitedAlgebraicGeometer on

Can sheafification be generalized to presheaves taking values in arbitrary categories (or at least a setting more general than presheaves of "nice" concrete categories)?

Comment #4484 by on

Dear ExcitedAlgebraicGeometer, yes. For example see Section 7.44 but there is a lot more you can do.

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