# Thom's first isotopy lemma

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 $S$?

• I must admit, I'm struggling a bit with your notation. ... – user102126 Jul 5 '18 at 9:28
• The notation is clear. – Ben McKay Jul 5 '18 at 9:44

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.
• 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. – Paul Sep 4 '18 at 4:25