Topologies in the vicinity of Euclidean space

Question

Given a smooth function $f:\mathbf R^n\to \mathbf R^m$ with $0$ as a regular value, I define the $(n-m)$ dimensional smooth manifold $M_f:=f^{-1}(0)$.

Let $f_0(x_1,...,x_n):=(x_1,...,x_m)$; $M_{f_0}$ is then simply $\mathbf R^{n-m}$. My question is: for every smooth $(n-m)$ dimensional manifold $\Sigma$, does there exist a function $g$ and a number $\delta>0$ such that $\forall \epsilon\in(0,\delta)$, $M_{f_0+\epsilon g}$ is diffeomorphic to $\Sigma$?

If this were the case then it could be said that all smooth manifolds are in the 'vicinity' of Euclidean space.

An example of this is: let $\Sigma=\mathbf S^2$ and $f_{\epsilon}(x,y,z):=\epsilon(x^2+y^2+z^2-1)+x$. $M_{f_0}=\mathbf R^2$, while $M_{f_\epsilon}$ has the topology of $\Sigma$ for all $\epsilon$ sufficiently small.

Answer 1

Under reasonable assumptions about $\Sigma$ the answer is yes. For example if $\Sigma$ is smooth and compact $(n-m)$-dimensional submanifold of $\mathbb{R}^n$ and it has trivial normal bundle*, that follows from the tubular neighborhood theorem. According to this theorem a neighborhood $U\subset\mathbb{R}^n$ of $\Sigma$ is diffeomorphic to $B^m\times\Sigma$. If $F:U\to B^m\times\Sigma$ is the diffeomorphism, $\pi:B^m\times\Sigma\to B^m$ is the projection onto the first factor, and $f=\pi\circ F$, then for every $y\in B^m$, $f^{-1}(y)$ is diffeomorphic to $\Sigma$.

This should also be true for smooth non-compact submanifolds with trivial normal bundle, because you can make tubular neighborhood gradually decrease as you move away. However, I did not think about that case carefully.

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*There was a mistake in my previous answer and as observed by Andy Putman in his comment.