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Let $X,Y$ be compact real analytic varieties. Suppose $Y$ is connected and there is a surjective analytic map $f:X\to Y$ such that each fiber of $f$ is connected. How to prove that $X$ is connected as well?

I believe this is a standard result but was not able to locate a reference. (The statement would follow if one proves that there are triangulations of $X$ and $Y$ such that $f$ is simplicial but I don't know the reference for such a statement either...)

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    $\begingroup$ That follows from point-set topology. For every partition $X=X'\sqcup X''$ into disjoint open and closed subsets, the images $f(X')$ and $f(X'')$ are closed subsets (since $f$ is proper). Since all fibers are connected (hence nonempty), the two subsets $f(X')$ and $f(X'')$ are disjoint and cover $Y$. Since $Y$ is connected, precisely one of $f(X')$ or $f(X'')$ is empty, and thus also $X'$, resp. $X''$, is nonempty. $\endgroup$
    – Jason Starr
    Commented Aug 12, 2022 at 0:28
  • $\begingroup$ Dear Jason, many thanks for this answer, it's great that it is so simple! $\endgroup$
    – aglearner
    Commented Aug 14, 2022 at 12:42

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