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I was thinking whether it would be possible to develop a converse to the preimage theorem in differential topology and I found the following post:

When is a submanifold of $\mathbf R^n$ given by global equations?

However, I didn't fully understand the main steps described in the post, even for the particular case when $M$ has codimension $1$. Just in case, I'm rephrasing the claim (particular case) and main steps of the proof below:

Claim: Given a connected and compact $n$-manifold $X \subset \mathbb{R}^{n+1}$, there always exists some smooth function $f: \mathbb{R}^{n+1} \rightarrow \mathbb{R}$, such that $X=f^{-1}(0)$ and $0$ is a regular value of $f$.

Sketch of the proof in the post: The post suggests that we should take a tubular neighborhood (where can I find a formal definition of the term "tubular neighborhood"?) $U$ of $X$ in $\mathbb{R}^{n+1}$ and identify $U$ with $X \times \mathbb{R}$. This will help us define a function $f: U \rightarrow \mathbb{R}$ with the required properties. Any idea how to define such $f$ from $U$? Moreover, after the function $f$ is defined on $U$, the post says that $f$ can be extended to the entire space $\mathbb{R}^{n+1}$ by a constant function, as $X$ has codimension $1$. How might such extension be constructed? Does this have anything to do with the fact that $X$ is connected and compact?

Thanks in advance for any help/suggestion!

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    $\begingroup$ A tubular neighbourhood $U$ of $X$ is, essentially by definition, a neighbourhood diffeomorphic to $X \times \mathbb R$; Wikipedia has the formal definition. The construction of such a neighbourhood is a straightforward application of compactness. Then $f$ could be, say, $(x, t) \mapsto (\arctan(t))^2$, and you can extend by $(\pi/2)^2$ outside of $U$. $\endgroup$
    – LSpice
    May 28, 2022 at 4:01
  • $\begingroup$ Hi there, thank you so much for your clarification! However, I'm still confused with a few points: Firstly, why do we have that the tubular neighborhood is diffeomorphic to $X \times \mathbb{R}$? I think the standard Tubular Neighborhood Theorem only gives us that the open neighborhood $U$ of $X$ is diffeomorphic to some open neighborhood of $X$ in the normal bundle $N(X;\mathbb{R}^{k+1})$. Thus, I guess we can only say that $U$ is diffeomorphic to $X \times V$ ($V$ is some open set in $\mathbb{R}$). How does connectedness and compactness of $X$ help us show that here $V$ can be $\mathbb{R}$ $\endgroup$
    – geooranalysis
    May 31, 2022 at 23:39
  • $\begingroup$ Secondly, when you say that "extend by $(\frac{\pi}{2})^2$ outside by $U$", do you just mean that $f(x) \equiv (\frac{\pi}{2})^2 \ (\forall \ x \notin U)$ is set to be a constant function outside of $U$? (One thing I forgot to mention is that we don't require $f$ here to be smooth on the entire space $\mathbb{R}^{k+1}$) $\endgroup$
    – geooranalysis
    May 31, 2022 at 23:39
  • $\begingroup$ Thirdly, I think another issue in your example is that the derivative of $(\arctan(t))^2$ is zero when $t=0$. Then the differential of the map is not necessarily surjective at $(x,0)$, i.e $0$ is not a regular value. I guess we can easily fix this by replacing $(\arctan(t))^2$ with $\arctan(t)$? $\endgroup$
    – geooranalysis
    May 31, 2022 at 23:42
  • $\begingroup$ Re, an open and connected subset of $\mathbb R$ is diffeomorphic to $\mathbb R$. $\endgroup$
    – LSpice
    Jun 1, 2022 at 14:39

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