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Let $X$ be a topological space and $A$ a set with the discrete topology. Consider the presheaf that associates with every open set $U$ of $X$ all the continuous maps $\mathcal{F}(U)$ of the form $U \rightarrow A$. If $U$ is connected, then $\mathcal{F}(U) \cong A$. If $U$ consists of $n$ disconnected components, then $\mathcal{F}(U)$ is isomorphic to the n-fold cartesian product of $A$ with itself. Is this a constant presheaf? Or do we have a constant presheaf only in the case where every open set of $X$ is connected, so that $\mathcal{F}(U) \cong A$?

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    I see your point. Let me try to fix this.2012-09-14

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The constant presheaf assigns the same set $A$ to every open set $U$ (including $\emptyset$). That is, it is built from constant maps to the set $A$, not from continuous maps to the discrete topological space $A$.

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    @Hagen: Surely you mean "and in this context $n^0 = 1$ and $0^n = 0$ for $n \neq 0$ and $0^0 = 1$".2013-04-26