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I'm currently reading a paper by X. Caicedo containing an introduction to sheaves. On page 8 he claims, that for every sheaf of sets $p:E\to X$ and every section $\sigma:U\to E$ (U being open in X) the image $\sigma(U)$ is open. This statement is proved by picking a point $e\in\sigma(U)$, an open neighborhood S of e, which satisfies

  1. $p(S)$ is open in X,
  2. $p\restriction S$ is a homeomorphism

and arriving at an open set $\sigma(U)\supseteq S\cap\sigma(U)=p^{-1}(p(S)\cap U)$.

I think the "$\supseteq$" part of this equation does not hold, if for example E is equipped with the discrete topology and the stalk of $p(e)$ has more than one element.

I have tried to show that $(p\restriction S)^{-1}(p(S)\cap U) = p^{-1}(U)\cap S$ is contained in $\sigma(U)$, but all attempts at that felt quite clumsy, leading me to believe I have missed something important about the structure of a sheaf.

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1 Answer

The inclusion $\sigma(U)\supseteq S\cap\sigma(U)$ is trivially true.
The equality $S\cap\sigma(U)=p^{-1}(p(S)\cap U)$ is false in general.
For a counterexample take any $X$ and $$E=X\times \lbrace 1,2\rbrace \:,p(x,i)=x\:, U=X\:, \sigma (x)=(x,1)\:, S=X\times \lbrace 1\rbrace$$ where $\lbrace 1,2\rbrace$ has the discrete topology.
You then have $S\cap\sigma(U)=S\neq p^{-1}(p(S)\cap U)=p^{-1}(X)=E $

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That's similar to what I had in mind. Thanks! –  pseudo-nym Jul 3 '12 at 13:25
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