Lemma 6.11.1. Let $\mathcal{F}$ be a sheaf of sets on the topological space $X$. For every open $U \subset X$ the map

is injective.

Let $X$ be a topological space. Let $x \in X$ be a point. Let $\mathcal{F}$ be a presheaf of sets on $X$. The *stalk of $\mathcal{F}$ at $x$* is the set

\[ \mathcal{F}_ x = \mathop{\mathrm{colim}}\nolimits _{x\in U} \mathcal{F}(U) \]

where the colimit is over the set of open neighbourhoods $U$ of $x$ in $X$. The set of open neighbourhoods is partially ordered by (reverse) inclusion: We say $U \geq U' \Leftrightarrow U \subset U'$. The transition maps in the system are given by the restriction maps of $\mathcal{F}$. See Categories, Section 4.21 for notation and terminology regarding (co)limits over systems. Note that the colimit is a directed colimit. Thus it is easy to describe $\mathcal{F}_ x$. Namely,

\[ \mathcal{F}_ x = \{ (U, s) \mid x\in U, s\in \mathcal{F}(U) \} /\sim \]

with equivalence relation given by $(U, s) \sim (U', s')$ if and only if there exists an open $U'' \subset U \cap U'$ with $x \in U''$ and $s|_{U''} = s'|_{U''}$. By abuse of notation we will often denote $(U, s)$, $s_ x$, or even $s$ the corresponding element in $\mathcal{F}_ x$. Also we will say $s = s'$ in $\mathcal{F}_ x$ for two local sections of $\mathcal{F}$ defined in an open neighbourhood of $x$ to denote that they have the same image in $\mathcal{F}_ x$.

An obvious consequence of this definition is that for any open $U \subset X$ there is a canonical map

\[ \mathcal{F}(U) \longrightarrow \prod \nolimits _{x \in U} \mathcal{F}_ x \]

defined by $s \mapsto \prod _{x \in U} (U, s)$. Think about it!

Lemma 6.11.1. Let $\mathcal{F}$ be a sheaf of sets on the topological space $X$. For every open $U \subset X$ the map

\[ \mathcal{F}(U) \longrightarrow \prod \nolimits _{x \in U} \mathcal{F}_ x \]

is injective.

**Proof.**
Suppose that $s, s' \in \mathcal{F}(U)$ map to the same element in every stalk $\mathcal{F}_ x$ for all $x \in U$. This means that for every $x \in U$, there exists an open $V^ x \subset U$, $x \in V^ x$ such that $s|_{V^ x} = s'|_{V^ x}$. But then $U = \bigcup _{x \in U} V^ x$ is an open covering. Thus by the uniqueness in the sheaf condition we see that $s = s'$.
$\square$

Definition 6.11.2. Let $X$ be a topological space. A presheaf of sets $\mathcal{F}$ on $X$ is *separated* if for every open $U \subset X$ the map $\mathcal{F}(U) \to \prod _{x \in U} \mathcal{F}_ x$ is injective.

Another observation is that the construction of the stalk $\mathcal{F}_ x$ is functorial in the presheaf $\mathcal{F}$. In other words, it gives a functor

\[ \textit{PSh}(X) \longrightarrow \textit{Sets}, \ \mathcal{F} \longmapsto \mathcal{F}_ x. \]

This functor is called the *stalk functor*. Namely, if $\varphi : \mathcal{F} \to \mathcal{G}$ is a morphism of presheaves, then we define $\varphi _ x : \mathcal{F}_ x \to \mathcal{G}_ x$ by the rule $(U, s) \mapsto (U, \varphi (s))$. To see that this works we have to check that if $(U, s) = (U', s')$ in $\mathcal{F}_ x$ then also $(U, \varphi (s)) = (U', \varphi (s'))$ in $\mathcal{G}_ x$. This is clear since $\varphi $ is compatible with the restriction mappings.

Example 6.11.3. Let $X$ be a topological space. Let $A$ be a set. Denote temporarily $A_ p$ the constant presheaf with value $A$ ($p$ for presheaf – not for point). There is a canonical map of presheaves $A_ p \to \underline{A}$ into the constant sheaf with value $A$. For every point we have canonical bijections $A = (A_ p)_ x = \underline{A}_ x$, where the second map is induced by functoriality from the map $A_ p \to \underline{A}$.

Example 6.11.4. Suppose $X = \mathbf{R}^ n$ with the Euclidean topology. Consider the presheaf of $\mathcal{C}^\infty $ functions on $X$, denoted $\mathcal{C}^\infty _{\mathbf{R}^ n}$. In other words, $\mathcal{C}^\infty _{\mathbf{R}^ n}(U)$ is the set of $\mathcal{C}^\infty $-functions $f : U \to \mathbf{R}$. As in Example 6.7.3 it is easy to show that this is a sheaf. In fact it is a sheaf of $\mathbf{R}$-vector spaces.

Next, let $x \in X = \mathbf{R}^ n$ be a point. How do we think of an element in the stalk $\mathcal{C}^\infty _{\mathbf{R}^ n, x}$? Such an element is given by a $\mathcal{C}^\infty $-function $f$ whose domain contains $x$. And a pair of such functions $f$, $g$ determine the same element of the stalk if they agree in a neighbourhood of $x$. In other words, an element if $\mathcal{C}^\infty _{\mathbf{R}^ n, x}$ is the same thing as what is sometimes called a *germ of a $\mathcal{C}^\infty $-function at $x$*.

Example 6.11.5. Let $X$ be a topological space. Let $A_ x$ be a set for each $x \in X$. Consider the sheaf $\mathcal{F} : U \mapsto \prod _{x\in U} A_ x$ of Example 6.7.5. We would just like to point out here that the stalk $\mathcal{F}_ x$ of $\mathcal{F}$ at $x$ is in general *not* equal to the set $A_ x$. Of course there is a map $\mathcal{F}_ x \to A_ x$, but that is in general the best you can say. For example, suppose $x = \mathop{\mathrm{lim}}\nolimits x_ n$ with $x_ n \not= x_ m$ for all $n \not= m$ and suppose that $A_ y = \{ 0, 1\} $ for all $y \in X$. Then $\mathcal{F}_ x$ maps onto the (infinite) set of tails of sequences of $0$s and $1$s. Namely, every open neighbourhood of $x$ contains almost all of the $x_ n$. On the other hand, if every neighbourhood of $x$ contains a point $y$ such that $A_ y = \emptyset $, then $\mathcal{F}_ x = \emptyset $.

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## Comments (2)

Comment #281 by Bas Edixhoven on

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