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.
Best Answer
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