# Definition of Constant Presheaf

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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Your sentence «For any open set $U$ of $X$ consider the presheaf thatassociates with $U$ all continuous maps $F(U)$ of the form $I\to A$» has a quantifier («For any open set $U$ of $U$») in the wrong place... Also: you mean «$U$ consists of $n$ connected components». –  Mariano Suárez-Alvarez Sep 14 '12 at 22:22
I see your point. Let me try to fix this. –  Manos Sep 14 '12 at 22:25

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 $A$.
By the way, $\mathcal{F}(\emptyset) = \emptyset$, right? (since there are no maps from $\emptyset$ to $A$). But does this not contradict the requirement $\mathcal{F}(\emptyset)=A$? –  Manos Sep 14 '12 at 22:31
@Manos That's not how the constant presheaf is defined. $F(\emptyset) = A$ (despite what Hartshorne might say). –  Zhen Lin Sep 15 '12 at 0:59
@Manos: Of course there is a map $\emptyset\to A$. You need to assign a value to every element of $\emptyset$, a task you have completed before even beginning. On the other hand, there is no map $A\to\emptyset$ unless $A=\emptyset$. The number of maps $A\to B$ is $|B|^{|A|}$ and in this context $n^0=0$ and $0^n=1$ for $n\ne0$ and $0^0=1$ (despite what they say in analysis). –  Hagen von Eitzen Sep 15 '12 at 9:39
@Hagen: Surely you mean "and in this context $n^0 = 1$ and $0^n = 0$ for $n \neq 0$ and $0^0 = 1$". –  dx7hymcxnpq Apr 26 '13 at 20:13