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Example 1.0.3. Let $X$ be a topological space, and A an abelian group. We define the constant sheaf $\mathscr{A}$ on $X$ determined by $A$ as follows. Give $A$ the discrete topology, and for any open set $U \subset X$, let $\mathscr{A}(U)$ be the group of all continuous maps of U into $A$. Then with the usual restriction maps, we obtain a sheaf $\mathscr{A}$. Note that for every connected open set $U$, $\mathscr{A}(U) \cong A$, whence the name "constant sheaf." If $U$ is an open set whose connected components are open (which is always true on a locally connected topological space), then $\mathscr{A}(U)$ is a direct product of copies of $A$, one for each connected component of $U$.

EXERCISE 1.1. Let $A$ be an abelian group, and define the constant presheaf associated to $A$ on the topological space $X$ to be the presheaf $U\longrightarrow A$ for all $U \not= \emptyset$, with restriction maps the identity. Show that the constant sheaf $\mathscr{A}$ defined in the text is the sheaf associated to this presheaf.

In reading this example I found some difficulties, which are:

In this example, it is said that $\mathscr{A}(U) \cong A$, for every connected open set $U$.

Well, I was not able to justify that. Is this justified by the discrete topology that is given to $ A $?

Following the example ... it says that: $\mathscr{A}(U)= \text{direct product of copies of $A$, one for each connected component of $U$}$.

Question: In that case, for $\mathscr{A}(U)$ to make sense, we should have that $\mathscr{A}(U)=\underbrace{A\times \cdots\times A}_{\text{finite number}}$? But for that to happen, it is necessary that $ U $ has a finite number of connected components. Why does $ U $ have a finite number of connected components?

In exercise 1.1, page 65, connected to this example, the solution I can think of depends on the fact: $\mathscr{A}(U)=\underbrace{A\times \cdots\times A}_{\text{finite number}}$.

Manoel
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    You need to make the question self-contained. Not everyone has Hartshorne on them. If this is the constant sheaf note that it's true that in fact $\mathscr{A}\left(\bigsqcup_i U)i\right)=\prod_i \mathscr{A}(U_i)$, so it's not that you have finitely many connected components, but that your connected components are clopen. This holds true if your scheme is locally Noetherian, etc. If not then you need to be more careful. For example, see this answer: https://math.stackexchange.com/questions/3178359/constant-sheaves-on-the-%C3%A9tale-site-of-a-scheme/ – Alex Youcis Sep 12 '20 at 19:31
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    Pictures of text are discouraged on this website. Please take the time to rewrite those images yourself, and keep this in mind for the future. – KReiser Sep 12 '20 at 21:01

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The basic idea here is that if $U$ is connected open, then $\mathscr{A}(U)$ is continuous maps of $U$ into $A$ with the discrete topology (as in the statement). The continuous image of a connected space is connected. Hence, it follows that any such function $f:U\to A$ must be constant and as such is uniquely determined by a choice of element in $A$. So, $\mathscr{A}(U)\cong A$ by sending $f\mapsto f(x)$ for any $x\in U$.

In the situation where $U=\bigsqcup_{i\in I} U_i$ with each $U_i$ open and hence connected, a continuous map $f:U\to A$ must now be constant on each $U_i$. In particular specifying a continuous map $f:U\to A$ is equivalent to specifying a sequence of elements of $A$, $(a_i)_{i\in I}$ such that $f(U_i)=\{a_i\}$ for each $i$. So, $\mathscr{A}(U)\cong \prod_{i\in I} A$.

Manoel
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Alekos Robotis
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