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According to wikipedia (constant sheaf) the constant sheaf $\underline{S}$ for an object $S$ is given by defining $\underline{S}(U)$ to be the set of functions $U \to S$, which are constant on each connected component of $U$. I believe this is only true if $S$ is trivial (terminal object) or the space $X$ is locally connected. For example if $X$ is totally disconnected, then $\underline{S}(U)$ consists of all functions $U \to S$, and the canonical map $\underline{S}$ $_x \to S$ won't be injective. I think that we have to define $\underline{S}(U)$ to be the set of locally constant functions $U \to S$. That is, continuous functions $U \to S$, where $S$ is endowed with the discrete topology. Since continuity is a local condition, this is a sheaf and it can be checked that the stalks are all $S$. This is the constant sheaf. Remark that locally constant functions are constant on each connected component, but the converse is not true. Interesting, in the wikipedia article about sheaves you can find the right definition.

Am I right or did I miss something?

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    $\begingroup$ theonion.com/articles/factual-error-found-on-internet,102 $\endgroup$ May 1, 2010 at 11:01
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    $\begingroup$ @Kevin Your onion plus one xkcd xkcd.com/386 $\endgroup$ May 1, 2010 at 11:33
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    $\begingroup$ Books also contain false definitions of constant sheaves. For example in Wells's "Differential analysis on complex manifolds", Prentice-Hall (1973), you can read page 39 (page 38 of the third edition : Springer, GTM 65, 2007), Example 1.6: "Let $X$ be a topological space and let $G$ be an abelian group. The assignment $U \to G$ is a sheaf, called the constant sheaf" [ $U$ is a nonempty open set according to the context on page 38]. The word connected doesn't even appear. But it's a fine book all the same... $\endgroup$ May 1, 2010 at 12:40
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    $\begingroup$ lmgtfy.com/?q=vandalize+every+equation (Warning: many of the links contain content that may offend and is definitely not safe for work). As Vinge wrote, "It is not called the Net of a Million Lies for nothing." $\endgroup$ May 1, 2010 at 13:09
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    $\begingroup$ So I think it is clear by now that wikipedia is wrong, and that you are right. I don't think anyone is going to post an answer to this question beyond what is already in the comments. So shouldn't this question be closed? $\endgroup$ May 1, 2010 at 17:14

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