3
$\begingroup$

While reading GMTW - The homotopy type of the cobordism category (https://arxiv.org/abs/math/0605249, p. 16) I found the following passage:

Lef $f:W\to \mathbb{R}$ be the projection. Then $(\pi_2,f):W\to X\times \mathbb{R}$ is proper since we have assumed that $X$ is compact. For $n\gg d$ we get an embedding $W\subset X\times \mathbb R\times \mathbb R^{d-1+n}$ which lifts $(\pi_2,f)$.

In this context I think $(\pi_2,f):W\to X\times \mathbb{R}$ is a submersion so by Ehresmann's lemma it is actually a fiber bundle. I'm not sure where the embedding comes from though. Is this some version of Whitney's theorem that is good for fiber bundles?

$\endgroup$
2
$\begingroup$

I think it is simpler than that if $W$ is assumed to be compact. For $n$ enough large, you have an embedding $\iota:W\to\mathbb{R}^{d-1+n}$ by Whitney's theorem assuming that $W$ is compact. A map such as $((\pi_2,f),\iota):W\to X\times\mathbb{R}\times\mathbb{R}^{d-1+n}$ is an example of the sort claimed in the above assertion. If $W$ is not necessarily compact, but you know that the fibre of $W\to X\times\mathbb{R}$ is compact, then you can apply a similar construction in a fibre-wise manner, I think!

$\endgroup$
  • $\begingroup$ You are right, thank you. Just to make sure, what's the problem with using Whitney's theorem if $W$ is not compact? $\endgroup$ – user109300 Sep 4 '17 at 7:23

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.