The constant sheaf has a sometimes mysterious definition but a nice equivalent description. I will explain it here below.

Let $X$ be a space and let $S$ be a set. Denote the $\operatorname{const}_{p,X}^S$ to constant presheaf with values in $S$, i.e., the presheaf $U\subset X\mapsto S$, with restriction maps all equal to $\operatorname{id}_S$); and denote $\operatorname{const}_{X}^S$ to the sheafification of $\operatorname{const}_{p,X}^S$. Note that the presheaf $\operatorname{const}_{p,X}^S$ is isomorphic to the presheaf of $S$-valued constant functions on $X$.

Recall that if $X$ is a space and $S$ is a set, then a function $f:X\to S$ is said to be locally constant if for every $x\in X$ there is a neighborhood $U\subset X$ of $x$ such that $f|_U$ is constant. Equivalently, $f$ is continuous when $S$ is endowed with the discrete topology. Let $\operatorname{LC}_X^S$ be the sheaf of locally constant $S$-valued functions on $X$. Note that, for $U\subset X$ open, $\operatorname{LC}_X^S|_U=\operatorname{LC}_U^S$. There is an obvious map $\operatorname{const}_{p,X}^S\to\operatorname{LC}_X^S$, which sends a section $s\in S$ of $\operatorname{const}_{p,X}^S$ over $U$ to the function defined at $U$ which is constantly equal to $s$. By the universal property of the sheafification gives us a map $\operatorname{const}_{X}^S\to\operatorname{LC}_X^S$.

The following result gives us a nice description of the constant sheaf:

Lemma. The canonical map $$ \label{map}\tag{1} \operatorname{const}_{X}^S\to\operatorname{LC}_X^S $$ is an isomorphism.

Proof. Fix $x\in X$.

We see that the induced map on stalks by \eqref{map} at $x\in X$ is bijective. Recall that the stalk of the constant (pre)sheaf is $(\operatorname{const}_X^S)_x=(\operatorname{const}_{p,X}^S)_x\cong S$.

For injectivity, note that the map induced by \eqref{map} on stalks at $x\in X$ sends the germ $s_x\in S$ to the germ of a function defined on an open neighborhood of $x$ which is constantly equal to $s_x$. This map is clearly injective.

For surjectivity, consider the germ of a locally constant function defined at an open neighborhood $U\subset X$ of $x$. Since the function is locally constant, we can shrink $U$ if necessary and assume that the function is actually constant in $U$. Let $s_x\in S$ be the value of this function. Now the germ $s_x$ of the stalk $(\operatorname{const}_X^S)_x$ is a pre-image through $\varphi_x$. $\square$

Applying the result “a locally constant function over a connected set is constant” we obtain a closed formula for the sections of the constant sheaf.

Since defining a locally constant function on a locally connected topological space amounts to giving values on each connected component of the space, we get

Corollary. Let $X$ a locally connected space and let $S$ a set. Then $$ \operatorname{const}_X^S(U)\cong\prod_{\operatorname{CC}(U)}S $$ for $U\subset X$ open, where $\operatorname{CC}(U)$ is the set of connected components of $U$.

I was wondering if we had some similar results with the extension by zero of the constant sheaf. I will introduce now the notations for the extension by zero. Fix an open set $U\subset X$ and let $j:U\to X$ be the inclusion. Given a presheaf $\mathcal{F}$ on $U$ of abelian groups, define the presheaf $j_{p!}\mathcal{F}$, the extension by zero presheaf, to be the presheaf on $X$ whose sections over $V\subset X$ are $$ j_{p !} \mathcal{F}(V)=\left\{\begin{array}{ccc}0 & \text { if } & V \not \subset U \\ \mathcal{F}(V) & \text { if } & V \subset U\end{array}\right. $$ with obvious restrictions mappings.

This presheaf has the stalk at $x\in X$ equal to $$ (j_{p!} \mathcal{F})_{x}=\left\{\begin{array}{cll}0 & \text { if } & x \notin U \\ \mathcal{F}_{x} & \text { if } & x \in U.\end{array}\right. $$

Define $j_!\mathcal{F}$, the extension by zero sheaf, to be the sheafification of $j_{p!}\mathcal{F}$. It can be shown that $j_!\mathcal{F}\cong j_!(\mathcal{F}^\#)$, where $\mathcal{F}^\#$ denotes the sheafification of $\mathcal{F}$. For that, consider the map $j_!\mathcal{F}\to j_!(\mathcal{F}^\#)$, the image of the canonical map $\mathcal{F}\to\mathcal{F}^\#$ under the functor $j_!:\operatorname{PSh}_\mathsf{Ab}(U)\to\operatorname{Sh}_\mathsf{Ab}(X)$, i.e., it is the image of $\mathcal{F}\to\mathcal{F}^\#$ first under the functor $j_{p!}:\operatorname{PSh}_\mathsf{Ab}(U)\to\operatorname{PSh}_\mathsf{Ab}(X)$ and then under the functor $(-)^\#:\operatorname{PSh}_\mathsf{Ab}(X)\to\operatorname{Sh}_\mathsf{Ab}(X)$. The induced maps on stalks by this sheaf morphism can be verified to be an isomorphism.

My question is: given an abelian group $A$, does the sheaf $j_!(\operatorname{const}_U^A)$ have some nice description in a similar spirit of the last lemma? Like “it is isomorphic to the sheaf of locally constant $A$-valued functions which vanish outside $U$” or something like that.

  • $\begingroup$ I am not sure about the validity of your Corollary: Take the space $X:=\mathbb{Q}$ equipped with the subspace topology coming from $\mathbb{R}$, and $S:=\mathbb{Z}$. The connected components of $X$ are all singletons, but a locally constant function $X\to S$ has to be constant in some open neighbourhood of each point $x\in X$, and singleton sets in $\mathbb{Q}$ are not open. $\endgroup$ Jul 15 at 7:20
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    $\begingroup$ @SimonWeinzierl Thank you. You are right. I've just edited the question, adding the hypothesis of local connectedness necessary to make the corollary work. I've done the same in the corollary of the answer. $\endgroup$ Jul 15 at 14:35

1 Answer 1


Yes, that is indeed the case. Denote $\operatorname{LC}_{X,U}^A$ to the sheaf of locally constant $A$-valued functions that vanish in $X\setminus U$ (convince yourself that this indeed is a sheaf). With this notation we have $\operatorname{LC}_{X,\varnothing}^A=\operatorname{LC}_{X}^A$. The sheaf $\operatorname{LC}_{X,U}^A$ is a subsheaf of $\operatorname{LC}_X^A$. There is an obvious map $\varphi:j_{p!}(\operatorname{const}_{p,U}^A)\to\operatorname{LC}_{X,U}^A$, which sends a section $a\in [j_{p!}(\operatorname{const}_{p,U}^A)](V)\subset A$ to the function defined on $V$ constantly equal to $a$ (check it is well-defined). Since $j_{!}(\operatorname{const}_{p,U}^A)=j_!(\operatorname{const}_{U}^A)$, we get a map $j_!(\operatorname{const}_{U}^A)\to\operatorname{LC}_{X,U}^A$. It suffices to verify that the induced map on stalks at $x\in X$ is an isomorphism.

The map $\varphi_x$ sends a germ $a_x\in A$ of $j_!(\operatorname{const}_{U}^A)_x$ to the function defined at an open neighborhood of $x$ with constant value $a_x$. Therefore $\varphi_x$ is injective. We see it is onto.

Case $x\notin U$. Then $j_!(\operatorname{const}_{U}^A)_x=0$. But also $(\operatorname{LC}_{X,U}^A)_x=0$, since for any locally constant function defined at an open neighborhood of $x$ and which vanishes at $x$ there is a subneighborhood where it is constantly zero.

Case $x\in U$. Consider the germ of a locally constant $A$-valued function defined at an open neighborhood $V$ of $x$. Shrinking the neighborhood if necessary, we can assume $V\subset U$, and also that the function is constant at $V$. Let $a\in A$ be the value of the function at $V$. Then the germ of $a\in[j_{p!}(\operatorname{const}_{p,U}^A)](V)=A$ is a pre-image under $\varphi_x$ of the germ of the function.

How does the isomorphism $j_!(\operatorname{const}_{U}^A)\to\operatorname{LC}_{X,U}^A$ acts on sections? Recall that given a presheaf $\mathcal{F}$ on $X$, a section over $V\subset X$ of the sheafification of $\mathcal{F}$ is given by a collection of compatible germs $(s_x)_{x\in V}$, where $s_x\in\mathcal{F}_x$ (see for example 007X). The isomorphism $j_!(\operatorname{const}_{U}^A)\to\operatorname{LC}_{X,U}^A$ maps the section $(a_x)_{x\in V}$ of $j_!(\operatorname{const}_{U}^A)$ over $V$ to the function \begin{align*} f:V&\to A\\ x&\mapsto a_x, \end{align*} which is locally constant and vanishes at $V\setminus U$, and therefore it is a section of $\operatorname{LC}_{X,U}^A$ over $V$.

Analogously as in the question, we have the following

Corollary. Let $X$ be a locally connected space, $U\subset X$ be an open set and $A$ be an abelian group. Then, for $V\subset X$ open, $$ j_!(\operatorname{const}_{X,U}^A)(V)\cong\prod_{D\in\operatorname{CC}(V)\text{ such that }D\subset U}A. $$


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