Let $X$ be a topological space and let $A$ be any set. In the literature, there seem to appear (at least) three different definitions for the constant sheaf $\mathcal{F}$ on $X$ associated to $A$, namely:
- $\mathcal{F}$ is the sheafification of the constant presheaf $F$ associated to $A$, which in turn is defined by $F(U) = A$ for every open $\emptyset \subsetneq U \subseteq X$, and $F(\emptyset) = \emptyset$.
- For $U \subseteq X$ open, $\mathcal{F}(U)$ is the set of continuous functions $U \to A$, where $A$ is equipped with the discrete topology. In other words, $\mathcal{F}(U)$ is the set of locally constant functions $U \to A$.
- $\mathcal{F}$ is the (unique?) sheaf of sets on $X$ such that $\mathcal{F}_x = A$ for every $x \in X$.
I understand why 1. and 2. are equivalent and imply 3.: Let $\mathcal{F}$ be as in 2.. Since constant functions are locally constant, identifying $a \in F(U)=A$ with the constant function $a \colon U \to A$ yields a morphism $F \to \mathcal{F}$ of presheaves. Denoting by $F^{sh}$ the sheafification of $F$, we obtain a morphism $F^{sh} \to \mathcal{F}$ of sheaves by the universal property of sheafification. However, one can check that the induced maps on stalks are isomorphisms (indeed, $F_x = F^{sh}_x = A$ and $\mathcal{F}_x=A$ for every $x \in X$, implying 3.), proving that $F^{sh} \cong \mathcal{F}$.
However, I do not see why 3. implies 1. or 2. In other words, assuming that $\mathcal{F}$ is a sheaf on $X$ such that $\mathcal{F}_x = A$ for every $x \in X$, then I do not see how to prove that, e.g., $\mathcal{F}(U)$ consists of all the locally constant functions $U \to A$.
One possible approach: Since $\mathcal{F}$ is a sheaf, we know that for every $U \subseteq X$ open, $f \in \mathcal{F}(U)$ is uniquely determined by the images $f_x$ of $f$ under the maps $\mathcal{F}(U) \to \mathcal{F}_x = A$ for every $x \in U$. In this way, every $f \in \mathcal{F}(U)$ can be identified with a function $f \colon U \to A$ (mapping $x \in U$ to $f_x \in A$). However, I am unable to prove that this map actually is locally constant.
Any help is appreciated, and have a happy new year!
I think there is a confusion in the third definition. What does it mean that $\mathcal F_x = A$?
If it means that for every $x$, the stalk $\mathcal F_x$ is isomorphic to $A$ in the category of sets, then this certainly is not a correct definition of constant sheaf, because it simply requires that the cardinality of $\mathcal F_x$ is the same as the cardinality of $A$. Under this defintion, the sheaf $\mathcal F$ can be quite arbitrary.
In order to make sense of this definition, you need a way to identify different stalks, so that there is the compatibility $\mathcal F_x = A = \mathcal F_y$. This extra condition will turn out to be exactly the locally constant condition in the second definition.