## 6.7 Sheaves

In this section we explain the sheaf condition.

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

1. 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 $i \in I$.

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

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

Remark 6.7.2. There is always a bit of confusion as to whether it is necessary to say something about the set of sections of a sheaf over the empty set $\emptyset \subset X$. It is necessary, and we already did if you read the definition right. Namely, note that the empty set is covered by the empty open covering, and hence the “collection of sections $s_ i$” from the definition above actually form an element of the empty product which is the final object of the category the sheaf has values in. In other words, if you read the definition right you automatically deduce that $\mathcal{F}(\emptyset ) = \textit{a final object}$, which in the case of a sheaf of sets is a singleton. If you do not like this argument, then you can just require that $\mathcal{F}(\emptyset ) = \{ *\}$.

In particular, this condition will then ensure that if $U, V \subset X$ are open and disjoint then

$\mathcal{F}(U \cup V) = \mathcal{F}(U) \times \mathcal{F}(V).$

(Because the fibre product over a final object is a product.)

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.

Comment #7074 by olof on

in e.g. 6.7.5, that symbol "A_x" isn't mean "stalk",right?

Comment #7075 by olof on

s_i and s_j agree on F(U_i∩U_j) right?

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