Thom's first isotopy lemma says that given $f:M\to P$ a smooth map between smooth manifolds and a closed Whitney stratified subset $S$ of $M$, such that $f|_S:S\to P$ is proper and $f|_X:X\to P$ is a submersion for any stratum $X$ of $S$, then $f|_S:S\to f(S)$ is a locally trivial fibration. Does this imply that $f|_X:X\to f(X)$ is a locally trivial fibration, for any stratum $X$ of $S$?

  • $\begingroup$ I must admit, I'm struggling a bit with your notation. ... $\endgroup$
    – user102126
    Jul 5, 2018 at 9:28
  • 10
    $\begingroup$ The notation is clear. $\endgroup$
    – Ben McKay
    Jul 5, 2018 at 9:44

2 Answers 2


Yes. Notice that since $f|_S: S \to S$ is proper, so is $f|_X$ (because $X$ is closed). Furthermore, $f|_X$ is a submersion and is surjective onto its image. So by a direct application of Ehresmann's fibration lemma you get your result.

  • $\begingroup$ I can use Ehresmann's fibration theorem if X is compact... $\endgroup$
    – Roushan
    Sep 4, 2018 at 2:40
  • $\begingroup$ You don't need X to be compact to use Ehresmann's fibration theorem. You need $f|_X$ to be a submersion (which is by hypothesis) and you need $f|_X$ to be proper (have a look at the link in the answer). Equivalently, you need that $f|_{S}^{-1}(p) \cap X= f|_X^{-1}(p)$ is a compact for all $p$ in $f(X)$. But this is true because $f|_S$ is proper, so $f|_S^{-1}(p)$ is a compact in $S$ and the intersection of a compact set with a closed set is a compact set. $\endgroup$
    – Paul
    Sep 4, 2018 at 4:25

Also note that because any fibration is also a compactification of a subset of all strata, we can always divide any subsets of these strata into an infinite number of compactified stratified and homotopically self similar spaces. Also note that the isotopy from strata 1 to strata 2 is a homotopy as well.


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