Let $f: M \to N$ a submersion between two manifolds, and let $S\subset N$ a subset of N. Proof that $S$ is a regular submanifold of $N$ if and only if $f^{-1}(S)$ is a regular submanifold of M.
So far I have half of the problem. Using the transversality theorem, we see that if $S$ is a submanifold, then using that $f$ is a submersion, we see that $S$ is transverse to $f$, so the preimage of $S$ is a submanifold of M. Now, for the second part I tried to use coordinate charts and the constant rank theorem, but nothing seems to work.
Any help will be appreciated.
Best Answer
HINT: First choose local coordinates on $M$ and $N$ so that $f(x_1,\dots,x_m) = (x_1,\dots,x_n)$ ($n\le m$). (Fix points corresponding to the origin and choose appropriate open sets, but I'm not going to mess with those.) Suppose the $(m-k)$-dimensional submanifold $Z=f^{-1}(S)$ given in these coordinates by $g(x)=0$, $g\colon \Bbb R^m\to\Bbb R^k$, with $0$ a regular value of $g$. Note that $g^{-1}(0)$ contains the fibers of the projection $f$.
Now define $h\colon\Bbb R^n\to\Bbb R^k$ by $h(x_1,\dots,x_n) = g(x_1,\dots,x_n,0,\dots,0)$. Then check that $h^{-1}(0) = S$ and that $0$ is a regular value of $h$. (The crucial observation is that at points of $Z$, we have $\dfrac{\partial g}{\partial x_i} = 0$ for $i=n+1,\dots,m$.)