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Let $\mathscr{F}$ and $\mathscr{G}$ by sheaves over a topological space $X$. For convenience, say they are sheaves of abelian groups. To be precise, I am using the etale space definition of a sheaf.

Say that for each $x \in X$ we have a group homomorphism on stalks $\phi_x \colon \mathscr{F}_x \to \mathscr{G}_x$. Define the map $\phi \colon \mathscr{F} \to \mathscr{G}$ to be the collation of the $\phi_x$ maps.

I would like to show that the following condition implies $\phi$ is continuous.

Whenever $s \in \Gamma(U, \mathscr{F})$, the map $s' \colon U \to \mathscr{G}$ defined by $s'(x) = \phi(s(x))$ is a section, i.e. $s' \in \Gamma(U, \mathscr{G})$.

My problem is working with an arbitrary open set in $\mathscr{G}$. I know that sections are necessarily open maps, so $s'(U)$ will be open in $\mathscr{G}$, but then $\phi^{-1}(s'(U)) = s(U)$ is open in $\mathscr{F}$... but this $s'(U)$ is not arbitrary open set in $\mathscr{G}$, so this does not suffice for a continuity proof. It seems I need to use my local homeomorphism somehow? I think maybe the local homeomorphism will allow me to show that the set $\{s(U) | \, U \text{ open in } X \}$ forms a basis for the topology of $\mathscr{G}$? If so, then my argument above is sufficient, but how to show this?

Perhaps this does it?

Let $U$ be open in $\mathscr{G}$. For each $x \in U$ there is some open set $V_x$ containing $x$ such that $\pi$ restricted to $V_x$ is a homeomorphism onto some set $U_x$ open in $X$ containing $\pi_x$. Now see that $$U = \bigcup_{x \in U} (V_x \cap U)$$ and each set $W_x = V_x \cap U$ is open in $\mathscr{G}$ and since $W_x \subseteq V_x$, we can further restrict the homeomorphisms $\pi|_{V_x}$ down to $\pi|_{W_x}$ and it is still a homeomorphism. Homeomorphisms are open maps thus the image $\pi|_{W_x}(W_x)$ is some open set $U'_x$ in $X$, and the inverse of the homeomorphism is a section, say $s_x$, on $U'_x$ and since the homeomorphism is bijective we have $s(U'_x) = W_x$. Putting this altogether, for each $W_x$ we have some set $U'_x$ with a section $s_x$ so that $s_x(U'_x) = W_x$, but the arbitrary open set $U = \bigcup W_x = \bigcup s_x(U'_x)$.

With this, the observation made earlier actually is sufficient for continuity.

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  • $\begingroup$ I think that your proof at the bottom works, although you should explicitly state what $\pi$ is here. $\endgroup$ – Sam Streeter Oct 25 '18 at 8:30
  • $\begingroup$ Hey Sam, I caught a flaw in my proof today. The preimage is not equality as I claim, but only containment! Whoops. Needs more to justify why picking up some extra germs does not sacrifice openness. $\endgroup$ – Prince M Oct 25 '18 at 21:41
  • $\begingroup$ I was able to get it using. Point wise definition of topological continuity $\endgroup$ – Prince M Oct 29 '18 at 18:13

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