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Assume all spaces are topological manifolds. A branched cover is a continuous open map with discrete fibers. A finite branched cover is one with finite fibers.

Questions. Given closed map $X\to S$ with finite discrete fibers to a compact base, does it admit an open embedding $X\hookrightarrow Y$ into a finite branched cover $Y\to S$? If so, why? If not, what are some simple examples and obstructions?

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    $\begingroup$ Are you assuming everything is surjective? Or are you just taking a very liberal definition of branched cover? $\endgroup$
    – Kevin Casto
    Commented Feb 10, 2023 at 20:02
  • $\begingroup$ @KevinCasto you can assume the bundles are surjective. $\endgroup$
    – Arrow
    Commented Feb 12, 2023 at 20:29
  • $\begingroup$ Ok. One brief comment is that since your map is closed and has compact fibers, it is proper, so in particular $X$ is compact $\endgroup$
    – Kevin Casto
    Commented Feb 12, 2023 at 22:52
  • $\begingroup$ I think the basic thing you need to worry about is say $f(x) = x^3 - x$, $(-2,2) \to (-6,6)$. The image of an open set around a critical point won't be open (instead half-open). This doesn't happen for complex polynomials bc the "2-dimensionality" somehow prevents it. Can this example be embedded in a branched cover? $\endgroup$
    – Kevin Casto
    Commented Feb 14, 2023 at 10:31
  • $\begingroup$ If you are happy embedding $S$ as well as $X$ (such that the restriction to $X$ gets taken to $S$), I think it's possible that analytic maps can be "complexified" in the way that my example obviously can be, and then you can apply the open mapping theorem. But that seems pretty different from what you want. $\endgroup$
    – Kevin Casto
    Commented Feb 14, 2023 at 10:36

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