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.

## 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 functor^{1}

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. 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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