# The $0$-section of sheaf

I ran into a problem with a defintion in Complex Analysis as follows:

A sheaf $\mathscr S$ over a paracompact Hausdorff space $X$ with a map $f: \mathscr S \to X$ such that

(1) $f$ is surjective and local homeomorphism,
(2) $f^{-1}(x)$ is an abelian group for every $x \in X$,
(3) the addtion operation is continuous on $\mathscr S$ where defined.

Then is the $0$-section of $\mathscr S$ an open set of $\mathscr S$ ?

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It seems like this is a sheaf in this sense. –  Dylan Moreland Dec 23 '11 at 2:10
Can you give me a proof? –  Hezudao Dec 23 '11 at 14:35

Allow me to use my own notation. First, let us show the following: for a local homeomorphism $\pi : E \to B$, any continuous section $s : U \to E$ is also a local homeomorphism, when $U$ is open in $B$. Indeed, by definition, $\pi \circ s : U \to B$ is simply the inclusion $U \hookrightarrow B$, and for each $x$ in $U$, there is an open neighbourhood $V$ of $s(x)$ in $E$ such that $\pi |_V : V \to \pi (V)$ is a homeomorphism and $\pi (V)$ is open in $B$; replacing $V$ by $V \cap \pi^{-1}(U)$ if necessary, we may assume $s^{-1}(V) = \pi(V)$, and then $s |_{\pi (V)} : \pi(V) \to V$ will be a homeomorphism. Hence, $s$ is a local homeomorphism, and in particular its image is open in $E$.
Now, let $\pi : E \to B$ be a local homeomorphism, such that the fibres are abelian groups, and suppose the addition operation $a : E \mathbin{\times_B} E \to E$ is continuous. Let $p_1 : E \mathbin{\times_B} E \to E$ and $p_2 : E \mathbin{\times_B} E \to E$ be the projection maps. Let $s : U \to E$ be any continuous section. By the preceding lemma, $s(U)$ is open in $E$, so $a^{-1}(s(U))$ is open in $E \mathbin{\times_B} E$; but $p_1^{-1}(s(U))$ is also open, so $a^{-1}(s(U)) \cap p_1^{-1}(s(U))$ is also open, and $p_2$ is an open map, so the set $$p_2(a^{-1}(s(U)) \cap p_1^{-1}(s(U)))$$ is open in $E$. But this is readily seen to be the image of the zero section $U \to E$, since the fibres of $\pi : E \to B$ are abelian groups.