# Characterizing Morphism of Sheaves

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.

• I think that your proof at the bottom works, although you should explicitly state what $\pi$ is here. – Sam Streeter Oct 25 '18 at 8:30
• 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. – Prince M Oct 25 '18 at 21:41
• I was able to get it using. Point wise definition of topological continuity – Prince M Oct 29 '18 at 18:13