Lemma 6.17.1. The presheaf $\mathcal{F}^{\# }$ is a sheaf.

## 6.17 Sheafification

In this section we explain how to get the sheafification of a presheaf on a topological space. We will use stalks to describe the sheafification in this case. This is different from the general procedure described in Sites, Section 7.10, and perhaps somewhat easier to understand.

The basic construction is the following. Let $\mathcal{F}$ be a presheaf of sets on a topological space $X$. For every open $U \subset X$ we define

where $(*)$ is the property:

For every $u \in U$, there exists an open neighbourhood $u \in V \subset U$, and a section $\sigma \in \mathcal{F}(V)$ such that for all $v \in V$ we have $s_ v = (V, \sigma )$ in $\mathcal{F}_ v$.

Note that $(*)$ is a condition for each $u \in U$, and that given $u \in U$ the truth of this condition depends only on the values $s_ v$ for $v$ in any open neighbourhood of $u$. Thus it is clear that, if $V \subset U \subset X$ are open, the projection maps

maps elements of $\mathcal{F}^{\# }(U)$ into $\mathcal{F}^{\# }(V)$. In other words, we get the structure of a presheaf of sets on $\mathcal{F}^{\# }$.

Furthermore, the map $\mathcal{F}(U) \to \prod _{u \in U} \mathcal{F}_ u$ described in Section 6.11 clearly has image in $\mathcal{F}^{\# }(U)$. In addition, if $V \subset U \subset X$ are open then we have the following commutative diagram

where the vertical maps are induced from the restriction mappings. Thus we see that there is a canonical morphism of presheaves $\mathcal{F} \to \mathcal{F}^{\# }$.

In Example 6.7.5 we saw that the rule $\Pi (\mathcal{F}) : U \mapsto \prod _{u\in U} \mathcal{F}_ u$ is a sheaf, with obvious restriction mappings. And by construction $\mathcal{F}^{\# }$ is a subpresheaf of this. In other words, we have morphisms of presheaves

In addition the rule that associates to $\mathcal{F}$ the sequence above is clearly functorial in the presheaf $\mathcal{F}$. This notation will be used in the proofs of the lemmas below.

**Proof.**
It is probably better for the reader to find their own explanation of this than to read the proof here. In fact the lemma is true for the same reason as why the presheaf of continuous function is a sheaf, see Example 6.7.3 (and this analogy can be made precise using the “espace étalé”).

Anyway, let $U = \bigcup U_ i$ be an open covering. Suppose that $s_ i = (s_{i, u})_{u \in U_ i} \in \mathcal{F}^{\# }(U_ i)$ such that $s_ i$ and $s_ j$ agree over $U_ i \cap U_ j$. Because $\Pi (\mathcal{F})$ is a sheaf, we find an element $s = (s_ u)_{u\in U}$ in $\prod _{u\in U} \mathcal{F}_ u$ restricting to $s_ i$ on $U_ i$. We have to check property $(*)$. Pick $u \in U$. Then $u \in U_ i$ for some $i$. Hence by $(*)$ for $s_ i$, there exists a $V$ open, $u \in V \subset U_ i$ and a $\sigma \in \mathcal{F}(V)$ such that $s_{i, v} = (V, \sigma )$ in $\mathcal{F}_ v$ for all $v \in V$. Since $s_{i, v} = s_ v$ we get $(*)$ for $s$. $\square$

Lemma 6.17.2. Let $X$ be a topological space. Let $\mathcal{F}$ be a presheaf of sets on $X$. Let $x \in X$. Then $\mathcal{F}_ x = \mathcal{F}^\# _ x$.

**Proof.**
The map $\mathcal{F}_ x \to \mathcal{F}^\# _ x$ is injective, since already the map $\mathcal{F}_ x \to \Pi (\mathcal{F})_ x$ is injective. Namely, there is a canonical map $\Pi (\mathcal{F})_ x \to \mathcal{F}_ x$ which is a left inverse to the map $\mathcal{F}_ x \to \Pi (\mathcal{F})_ x$, see Example 6.11.5. To show that it is surjective, suppose that $\overline{s} \in \mathcal{F}^\# _ x$. We can find an open neighbourhood $U$ of $x$ such that $\overline{s}$ is the equivalence class of $(U, s)$ with $s \in \mathcal{F}^\# (U)$. By definition, this means there exists an open neighbourhood $V \subset U$ of $x$ and a section $\sigma \in \mathcal{F}(V)$ such that $s|_ V$ is the image of $\sigma $ in $\Pi (\mathcal{F})(V)$. Clearly the class of $(V, \sigma )$ defines an element of $\mathcal{F}_ x$ mapping to $\overline{s}$.
$\square$

Lemma 6.17.3. Let $\mathcal{F}$ be a presheaf of sets on $X$. Any map $\mathcal{F} \to \mathcal{G}$ into a sheaf of sets factors uniquely as $\mathcal{F} \to \mathcal{F}^\# \to \mathcal{G}$.

**Proof.**
Clearly, there is a commutative diagram

So it suffices to prove that $\mathcal{G} = \mathcal{G}^\# $. To see this it suffices to prove, for every point $x \in X$ the map $\mathcal{G}_ x \to \mathcal{G}^\# _ x$ is bijective, by Lemma 6.16.1. And this is Lemma 6.17.2 above. $\square$

This lemma really says that there is an adjoint pair of functors: $i : \mathop{\mathit{Sh}}\nolimits (X) \to \textit{PSh}(X)$ (inclusion) and $\# : \textit{PSh}(X) \to \mathop{\mathit{Sh}}\nolimits (X)$ (sheafification). The formula is that

which says that sheafification is a left adjoint of the inclusion functor. See Categories, Section 4.24.

Example 6.17.4. See Example 6.11.3 for notation. The map $A_ p \to \underline{A}$ induces a map $A_ p^\# \to \underline{A}$. It is easy to see that this is an isomorphism. In words: The sheafification of the constant presheaf with value $A$ is the constant sheaf with value $A$.

Lemma 6.17.5. Let $X$ be a topological space. A presheaf $\mathcal{F}$ is separated (see Definition 6.11.2) if and only if the canonical map $\mathcal{F} \to \mathcal{F}^\# $ is injective.

**Proof.**
This is clear from the construction of $\mathcal{F}^\# $ in this section.
$\square$

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