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This is the second part of the following solved question.

[I'm following Bredon's Book]. After explaining the idea behind the "desired" bijection we want to build, Bredon start dealing with the well-definitedness of such association:

Now suppose we are given two fattened $k-$manifolds $g_0 \colon M^k_0\times E^n \to \mathbb{R}^{n+k}$ and $g_1\colon M^k_1 \times E^n \to \mathbb{R}^{n+k}$ and that the associated maps are homotopic: $\phi_{g_0} \simeq \phi_{g_1}$ via the homotopy $F \colon \mathbb{R}^{n+k} \times I \to \mathbb{R}^{n}_+$.

By composing $F$ with a map $\mathbb{R}^{n+k} \times I \to \mathbb{R}^{n+k}\times I$ of the form $1 \times \psi$ where $\psi(t)=0$ for $t$ near $0$ and $\psi(t)=1$ for $t$ near $1$, we can assume that $F$ is a constant homotopy near the two ends. Also, of course, $F$ can be assumed to be smooth away from $F^{-1}(\infty)$.

Let $q\in \mathbb{R}^n$ be a regular value of $F$ and put $V^{k+1}=F^{-1}(\{q\})$. Then there is an open disk $D^n$ about $q$ and an embedding $V^{k+1}\times D^n \to \mathbb{R}^{n+k}\times I$ onto a neighbourhood $W$ of $V$ and whose inverse is $r \times F \colon W \to V^{k+1} \times D^n$, $r$ being the normal retraction.

I don't understand how Bredon applies the theorem about the regular value because in the hypothesis he gave, it's required that the pre image of a regular point is compact. The last time he used this result it was clear that such pre-image is in fact compact (because we started with maps form the sphere-hence compact) but now the homotpy is defined on $\mathbb{R}^{n+k}\times I$ which is surely non-compact

Someone can explain why the result still holds here? (or why the highlighted part is true?) For notation please refer to my linked question above.

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The content of the regular value theorem is only a local statement (as being a submanifold is). Hence you don't need compactness at all to apply it.

Also note that the embedding $V^{k+1 }\times D^n \to R^{n+k}\times I$ exists by the regular neighborhood theorem and the fact that normal bundles pull back.

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    $\begingroup$ Note that I was just answering your question. In general we are only interested in compact cobordisms ($M \times [0,1)$ makes every $M$ nullbordant otherwise), so it makes sense to hope for the closed preimage to be also bounded. So this is why he constructs the maps and homtopies carefully to be only non-trivial on a compact set and the preimage of a point (which cannot be the basept) is a closed subset of this compact set. Being redundant: $F^{-1}(q)\subset F^{-1}(R^n_+ -\infty)$ are both by construction compact. $\endgroup$ – Daniel Valenzuela Aug 6 '15 at 15:07
  • $\begingroup$ Consider my answer as the correct answer to your question and my comment as the answer you want. $\endgroup$ – Daniel Valenzuela Aug 6 '15 at 15:09
  • $\begingroup$ Dear Dan, thank you for your answer, I got the point, accept it, but now I'm curious about the idea behind the existence of the embedding you mention. On Bredon there is (obviously) the formal proof, but I'm not able to grasp the actual reasoning behind it. Even though it's not completely related to my question, do you mind elaborate a little the last part of your answer? even as a comment :) Thanks anyway! $\endgroup$ – Riccardo Aug 6 '15 at 20:26
  • $\begingroup$ Sure, by the tubular neighborhood theorem there exists an embedding of the normal bundle $\nu V^{k+1} \to R^{n+k}\times I$ which extends the embedding of $V^{k+1}$ (the 0-section). Now we have that $\nu V^{k+1} = F|V^*(\nu \{q\}) = F|V^*(\{q\} \times R^n = V \times R^n$. $\endgroup$ – Daniel Valenzuela Aug 7 '15 at 6:20
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At the top of the page 119 there is an assumption that maps and homotopies $\mathbb R^{n+k}\to S^n$ are constant to the base point outside some compact subset. This is expected, when we are interested in mappings between spheres.

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  • $\begingroup$ thanks for pointing me out this detail! $\endgroup$ – Riccardo Aug 6 '15 at 20:26

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