Does there exist an orientable manifold $M$ and a closed orientable submanifold $Z \subseteq M$ with codimension one so that $Z$ is not the preimage of a regular value of a smooth function $f:M\to N$?
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Won't $Z$ have codimension $\dim N$? – Qiaochu Yuan Sep 10 '20 at 22:39
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Yes. I erased the requirement of codimension one. – Rasmus Sep 11 '20 at 10:40
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I rolled back your edit. Now the question has been answered, it is better not to change. – Arctic Char Sep 11 '20 at 10:59
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The general question is answered here – Arctic Char Sep 11 '20 at 12:50
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Let $T$ be a tubular neighbourhood of $Z$ in $M$. The normal bundle of $Z$ in $M$ is an orientable line bundle and is therefore trivial. So there is a diffeomorphism $\phi : T \to Z\times(-\pi,\pi)$ such that $\phi(z) = (z, 0)$ for $z \in Z$. Now consider the map $f_0 : M \to S^1$ given by
$$f_0(m) = \begin{cases} \exp(i\operatorname{pr}_2(\phi(m))) & m \in T\\ -1 & m \not\in T. \end{cases}$$
Note that $f_0|_{\phi^{-1}(Z\times[-1,1])}$ is smooth, so we can deform to a smooth map $f: M \to S^1$ which agrees with $f_0$ on $\phi^{-1}(Z\times[-1,1])$. Now note that $1 \in S^1$ is a regular value of $f$ with $f^{-1}(1) = Z$.
More generally, if $Z$ is a codimension $k$ submanifold of $M$ with trivial normal bundle, then $Z$ is the preimage of a regular value of a map $f : M \to S^k$. This idea was developed by Thom and leads to the notion of Thom spaces and the Pontryagin-Thom construction.
Michael Albanese
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