Section 6.7 (006S): Sheaves—The Stacks project

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 $\mathop{\mathit{Sh}}\nolimits (X)$ .

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.

Comments (8)

Comment #7074 by olof on February 22, 2022 at 10:57

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

Comment #7075 by olof on February 22, 2022 at 11:15

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

Comment #7253 by Johan on May 01, 2022 at 10:56

Yes and yes.

Comment #8595 by Paul Le Meur on July 25, 2023 at 06:56

[It seems to me that the following remark is correct, but i should write it out with even more care to be 100%. Sorry if there is any mistake.] In remark 6.7.2 the argument that -with the definition of sheaf given here- the image of the empty set must be a final object of the value category does not seem complete to me -and perhaps the conclusion even fails. We do have that « 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 ». But then we must justify that this yields a morphism from $\mathcal F(\emptyset)$ to this product (a final object of the value category) and that this morphism is an isomorphism -for instance because it is an equalizer of identity morphisms. Actually this requires some reinterpretation of the definition in terms of elements given here, because if $I=\emptyset$ there is no $s_i$ and the ensuing uniqueness condition is vacuous. For the argument to work we can reformulate the definition in terms of equalizer diagram. Alternatively we can -as done in Hartshorne, p61- formulate separately the locality axiom, this shows that the short "combination" of the locality axiom with the gluing axiom into a single axiom like here is not equivalent to the conjunction of the 2 standard axioms. The formulations look equivalent but the one given here is actually weaker because the empty family argument does not work: so a sheaf as defined here can have a nonfinal object as value on the empty set.

Comment #8596 by Johan on July 27, 2023 at 09:32

The point of 6.7.2 was to prevent having this discussion!

Comment #8597 by Paul Le Meur on July 27, 2023 at 11:42

Dear Aise Johan, Sorry. :)

But i must correct myself: i think that the axiom as stated here does imply that $\mathcal F(\emptyset)=0$ . The argument though has to be modified: we see that in "there exists a unique section $s\in\mathcal F(\emptyset)$ such that $s_i=s|_U_i$ for all $i\in I$ " the restriction condition is always true as $i\in I$ is always false, so the first part says that there exists a section $s\in\mathcal F(\emptyset)$ and that any section $t\in\mathcal F(\emptyset)$ (which necessarily satisfies the always-true restriction condition) equals $s$ , thus $\mathcal F(\emptyset)$ is indeed $0$ .

Also i should have said that Hartshorne actually imposes all presheaves to have $\mathcal F(\emptyset)=0$ , so sheaves a fortiori satisfy this, though his axiom for sheaves would of course enforce this, as above.

And to finish with the equalizer argument: category theorists and logicians surely know when it is possible to translate exactly (even with empty set subtleties) between formulas and diagrams. Here i think it is not trivial to watertight-prove that the set formulation of the axiom does translate into the expected diagrams (in all types of value categories), including all subtleties of vacuous conditions.

I am really sorry to have annoyed you -plus making mistakes in my comment. I can only imagine how tedious this project is -hopefully there are some funnier sections than this one. So i thank you for all your work here and elsewhere. Best wishes, Paul

Comment #9165 by Johan on May 13, 2024 at 15:20

No problem and I enjoy working on this!

Comment #10061 by Sinnaruil on March 24, 2025 at 20:32

I wonder could we add some categorical definition here? It seems convenient to define a sheaf as a special contravariant functor.

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6.7 Sheaves

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