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Ehresmann's theorem for manifolds states: If $f : X \to Y$ is a proper submersion, then $X$ is a locally trivial fibration on $Y$.

Some sources I am reading (Lazarsfeld Positivity in Algebraic Geometry, Book I, particularly lemma 3.3.2) suggest (but do not say outright) that if $f$ is in the category of $\mathbb{C}$ varieties, we can get by without properness, provided we are happy with removing some closed subsets from $X$ and $Y$. (The local triviality is still in the topological category of course.)

It seems (from googling) like this has something to do with Whitney stratifications, Nagata's theorem, and Thom's first isotopy theorem - I am looking at 5.1 in http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002092808&physid=PHYS_0318 and some other internet sources.

What I am thinking so far, with much uncertainty, is that you proceed something like:

Assume $f : X \to Y$ is a smooth morphism of smooth varieties. Using Nagata's theorem, factor $f$ as an open immersion followed by a proper morphism. $X \to \bar{X} \to Y$. Call the proper map $\bar{f}$.

Then $\bar{X}$ admits a Whitney stratification, and we know on the dense open $X$ that $f$ is a submersion. Then, by some continuity argument or magic or ????, we (maybe) also know that $f$ is a submersion when restricted to the other stratum of $\bar{X}$. Then we apply Thom's first isotopy theorem to conclude that $\bar{f}$ is Euclidean locally an isomorphism.

Finally, work with $f^{-1} ( \bar{f}(\bar{X} \setminus X) \to \bar{f}(\bar{X} \setminus X)$.

Can someone clarify the magic / ???? step above? Why is $\bar{f}$ a submersion on the other strata? Is it? What is the precise statement? I am wary of citing an old French source I don't understand, and I would prefer to have a clear picture of how this argument should proceed. I really don't have much experience with this field so ... any input would be good.

PROPOSITION 11.1. (Thom's first isotopy lemma.) Suppose $f | S : S → P$ is proper and $f | S : S → P$ is a submersion for each stratum X of S. Then the bundle $(S, f, P)$ is locally trivial.

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  • $\begingroup$ The argument you present here seems quite different from what Verdier does in the paper you cite. Verdier actually works with the open $Y \setminus \bar f(\bar X)$ (here the fibre is empty, so the result is trivially true), along with the interior of some good locally closed set above which he knows the result. To get this latter set in the first place, he refers to his earlier theorem 3.3. $\endgroup$
    – R. van Dobben de Bruyn
    Nov 28, 2016 at 5:08
  • $\begingroup$ It would help if you could specify a particular statement in his argument that you are confused by. $\endgroup$
    – R. van Dobben de Bruyn
    Nov 28, 2016 at 5:09
  • $\begingroup$ You can take a resolution of singularities of the variety $\bar X$, viewed over the field of functions of $Y$, and spread it out (by performing the blowing-ups of the closures of the loci you blow-up generically, for example). Then you have reduced to the case where $\bar X$ is smooth over a dense open subscheme of $Y$. $\endgroup$
    – ACL
    Nov 28, 2016 at 11:29

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