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 $s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j}$ 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

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 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (6)

Comment #2953 by Ko Aoki on

Comment #3080 by Johan on

Comment #3979 by Lucy on

Comment #4106 by Johan on

Comment #4332 by ExcitedAlgebraicGeometer on

Comment #4484 by Johan on