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.
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.