I'm getting very confused distinguishing the difference between a locally constant sheaf and a constant sheaf.
$\textbf{Constant sheaf}$: Let M be a vector space. The constant sheaf over $X$ is given by $$\underline{M}(U) = \{s: U \to M | s\text{ constant on connected components}\}$$ where $U\subset X$ open, with restriction given by restriction of functions gives the constant sheaf with value M.
$\textbf{Locally constant sheaf}$: A sheaf $\mathcal{F}$ such that there is an open cover ${U_{\alpha}}$ of X with $\mathcal{F}|_{U_{\alpha}}$ is a constant sheaf.
I know that a constant sheaf is a locally constant sheaf but I don't see why a locally constant sheaf isn't a constant sheaf.
Take the example that you have $X = U_1 \sqcup U_2$, the disjoint union of two connected open sets, and take two different constant sheaves defined on each connected piece. This is supposedly locally a constant sheaf but not a constant sheaf. But by definition of the constant sheaf, we have that this sheaf is constant on connected components as it is constant on $U_1$ and constant on $U_2$, no? But this is the same logic I would use for locally constant sheaf?
Also, if anyone could shed some light on how to view the covering map over the circle, $z \to z^2$, as a sheaf that would be great. How is this a locally constant sheaf? It's a map, not a sheaf?
I've seen this post, but I think what I want is it explained in more detail Locally constant sheaf but not constant
Let $A$ be a space endowed with a discrete topology, and $X$ a topological space, the constant sheaf defined by $A$ is the sheaf such that for every open subset $U$ of $X$, $M(U)$ is the space of continuous maps $U\rightarrow A$.
Let $X=\{x_1,x_2\}$, $U_1=\{x_1\},U_2=\{x_2\}$ are open subset Let $A_1=\{a\}, A_2=\{a,b\}$, defined $M(U_i)$ is the space of continuous map $U_i\rightarrow A_i$ and $M(X)$ is the space of continuous maps $X\rightarrow \{a,b\}$ $M$ is a sheaf if $s_i:U_i\rightarrow A_i$, we can extend $s_i$ to $X$ by setting $s(x_i)=s_i(x_i)$, remark that $s$ may not be constant but continuous since $X$ is not connected.
$M$ is locally constant since it is constant on $U_i$, but $M$ is not constant.