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Suppose that $f:X\to S$ is a holomorphic morphism of Hausdorff complex manifolds and that $s\in S$ such that $f^{-1}(s)$ is compact (and maybe singular). Then is it true that there is an open neighborhood $U$ of $s$ in $S$ such that $f$ is proper over $U$?

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No. Let $f$ be the inclusion of an open subset. –  Tom Goodwillie Nov 4 '11 at 10:56
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@Tom: if $f$ is an open immersion, the answer is yes (take $U=f(X)$). Inkspot doesn't ask $f$ to be proper over $S$. –  Qing Liu Nov 4 '11 at 21:34
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You could take a point in the boundary of $U$; then the inverse image is compact, since it is empty, but the function is not proper in a neighborhood of the point. –  Angelo Nov 4 '11 at 21:58
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@Inkspot: This is true under the additional (strong) hypothesis that $f$ is a submersion. The proof is by standard techniques in point set topology. –  Gunnar Magnusson Nov 5 '11 at 7:54
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@Gunnar, I guess you want to suppose the fibers of $f$ are connected. In my counterexample below, $f$ is a submersion. –  Qing Liu Nov 5 '11 at 13:04
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2 Answers 2

No. Let $g:Y\to S$ be an arbitrary finite morphism, $Z\subset Y$ a closed subset such that $Z$ does not contain an entire fibre (for instance any closed point works if the degree of $g$ is larger than $1$). Let $X=Y\setminus Z$ and $f=g|_X$. Then for any point $s\in S$, $f^{-1}(s)$ is compact (a set of finite points) and non-empty but for any point $s\in g(Z)$ there is no neighbourhood of $s$ in $S$ over which $f$ would be proper.

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Consider $S=\mathbb C^{\star}$, $X=\mathbb C^* \setminus \lbrace 1 \rbrace$ and $f(z)=z^2$. Then $f^{-1}(1)=\lbrace 1\rbrace $ is proper, but $f^{-1}(U)\to U$ is never proper for any open neighborhood $U$ of $1$.

However, if $X, Y$ and $f$ are algebraic, then the answer is yes under the hypothesis $f^{-1}(s)$ is non-empty, $f$ is flat (as $X, Y$ are smooth, this comes down to say that all fibers of $f$ are equidmensional of the same dimension), and the fibers of $f$ are connected. See EGA IV.15.7.10. The conclusion is $f^{-1}(U)\to U$ is proper over a Zariski open neighborhood $U$ of $s$. There is probably a holomorphic version of this result.

EGA III.5.5.2 can also be interesting (over the formal completion of $O_{S,s}$).

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