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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

Arrow
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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.

Tsemo Aristide
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  • Does that mean that on a disconnected set, we can never have a constant sheaf (as the gluing condition wouldn't be satisfied) but we can have a locally constant sheaf? – Arrow Apr 20 '20 at 01:39
  • If $X$ is any topological set, take $A$ a non empty set and define $M(U)=A$ if $U$ is an open subset, the restriction maps is the identity define a constant sheaf on $X$ – Tsemo Aristide Apr 20 '20 at 02:26
  • Take $X=U_1 \sqcup U_2$. By definition, $M(U_1)=\mathbb{Z}$, $M(U_2)=\mathbb{Z}$ and $M(X)=\mathbb{Z}$ as $X$ is also clearly open. Take $s_i \in M(U_i)$ such that $s_i=i$, so $s_1=1$ and $s_2=2$. I'm assuming the restriction maps to the empty set is just the trivial map, in which case, $s_1|{U_1 \cap U_2} = s_2|{U_1 \cap U_2}$, as $U_1 \cap U_2 = \emptyset$. But this doesn't glue to a section of $s \in M(X)$ that is constant, as we would require $s =k$ where $k \in \mathbb{Z}$? – Arrow Apr 20 '20 at 02:38
  • Constant sheaf means that $M(U)=A$ and the restrictions functions are the identity, it does not means that the sections are constant – Tsemo Aristide Apr 20 '20 at 02:43
  • I think this is the part I'm missing, any chance you could explain it in a bit more detail? I'm not seeing where that example I just gave is going wrong, am I incorrectly using the definitions? – Arrow Apr 20 '20 at 02:50
  • If $M(U)=\mathbb{Z}$, a section $s \in M(U)$ has to be constant $s=k$ for some $k\in \mathbb{Z}$, no? – Arrow Apr 20 '20 at 02:52
  • It is constant if $U$ is connected – Tsemo Aristide Apr 20 '20 at 03:04
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    @TsemoAristide Sorry but your definition of constant sheaf is wrong, you only define the constant presheaf. The definition in the OP is the good one. Moreover I don't understand how $M$ is a sheaf. You said $M(X)$ is the set of continuous maps $X\to{a,b}$. How do you restrict the constant map with value $b$ on $U_1$ if $M(U_1)$ is the space of continuous map $U_1\to{a}$ ? – Roland Apr 20 '20 at 07:02