## 6.7 Sheaves

In this section we explain the sheaf condition.

Definition 6.7.1. Let $X$ be a topological space.

A *sheaf $\mathcal{F}$ of sets on $X$* is a presheaf of sets which satisfies the following additional property: Given any open covering $U = \bigcup _{i \in I} U_ i$ and any collection of sections $s_ i \in \mathcal{F}(U_ i)$, $i \in I$ such that $\forall i, j\in I$

\[ s_ i|_{U_ i \cap U_ j} = s_ j|_{U_ i \cap U_ j} \]

there exists a unique section $s \in \mathcal{F}(U)$ such that $s_ i = s|_{U_ i}$ for all $i \in I$.

A *morphism of sheaves of sets* is simply a morphism of presheaves of sets.

The category of sheaves of sets on $X$ is denoted $\mathop{\mathit{Sh}}\nolimits (X)$.

Example 6.7.3. Let $X$, $Y$ be topological spaces. Consider the rule $\mathcal{F}$ which associates to the open $U \subset X$ the set

\[ \mathcal{F}(U) = \{ f : U \to Y \mid f \text{ is continuous}\} \]

with the obvious restriction mappings. We claim that $\mathcal{F}$ is a sheaf. To see this suppose that $U = \bigcup _{i\in I} U_ i$ is an open covering, and $f_ i \in \mathcal{F}(U_ i)$, $i\in I$ with $f_ i |_{U_ i \cap U_ j} = f_ j|_{U_ i \cap U_ j}$ for all $i, j \in I$. In this case define $f : U \to Y$ by setting $f(u)$ equal to the value of $f_ i(u)$ for any $i \in I$ such that $u \in U_ i$. This is well defined by assumption. Moreover, $f : U \to Y$ is a map such that its restriction to $U_ i$ agrees with the continuous map $f_ i$. Hence clearly $f$ is continuous!

We can use the result of the example to define constant sheaves. Namely, suppose that $A$ is a set. Endow $A$ with the discrete topology. Let $U \subset X$ be an open subset. Then we have

\[ \{ f : U \to A \mid f\text{ continuous}\} = \{ f : U \to A \mid f\text{ locally constant}\} . \]

Thus the rule which assigns to an open all locally constant maps into $A$ is a sheaf.

Definition 6.7.4. Let $X$ be a topological space. Let $A$ be a set. The *constant sheaf with value $A$* denoted $\underline{A}$, or $\underline{A}_ X$ is the sheaf that assigns to an open $U \subset X$ the set of all locally constant maps $U \to A$ with restriction mappings given by restrictions of functions.

Example 6.7.5. Let $X$ be a topological space. Let $(A_ x)_{x \in X}$ be a family of sets $A_ x$ indexed by points $x \in X$. We are going to construct a sheaf of sets $\Pi $ from this data. For $U \subset X$ open set

\[ \Pi (U) = \prod \nolimits _{x \in U} A_ x. \]

For $V \subset U \subset X$ open define a restriction mapping by the following rule: An element $s = (a_ x)_{x\in U} \in \Pi (U)$ restricts to $s|_ V = (a_ x)_{x \in V}$. It is obvious that this defines a presheaf of sets. We claim this is a sheaf. Namely, let $U = \bigcup U_ i$ be an open covering. Suppose that $s_ i \in \Pi (U_ i)$ are such that $s_ i$ and $s_ j$ agree over $U_ i \cap U_ j$. Write $s_ i = (a_{i, x})_{x\in U_ i}$. The compatibility condition implies that $a_{i, x} = a_{j, x}$ in the set $A_ x$ whenever $x \in U_ i \cap U_ j$. Hence there exists a unique element $s = (a_ x)_{x\in U}$ in $\Pi (U) = \prod _{x\in U} A_ x$ with the property that $a_ x = a_{i, x}$ whenever $x \in U_ i$ for some $i$. Of course this element $s$ has the property that $s|_{U_ i} = s_ i$ for all $i$.

Example 6.7.6. Let $X$ be a topological space. Suppose for each $x\in X$ we are given an abelian group $M_ x$. Consider the presheaf $\mathcal{F} : U \mapsto \bigoplus _{x \in U} M_ x$ defined in Example 6.4.5. This is not a sheaf in general. For example, if $X$ is an infinite set with the discrete topology, then the sheaf condition would imply that $\mathcal{F}(X) = \prod _{x\in X} \mathcal{F}(\{ x\} )$ but by definition we have $\mathcal{F}(X) = \bigoplus _{x \in X} M_ x = \bigoplus _{x \in X} \mathcal{F}(\{ x\} )$. And an infinite direct sum is in general different from an infinite direct product.

However, if $X$ is a topological space such that every open of $X$ is quasi-compact, then $\mathcal{F}$ *is* a sheaf. This is left as an exercise to the reader.

## Comments (3)

Comment #7074 by olof on

Comment #7075 by olof on

Comment #7253 by Johan on