In the book "Topology from the Differential Viewpoint" (Milnor) he proves on page 11 the following lemma:
If $f: M\to N$ is a smooth map between manifolds of dimension $m\geq n$ and if $y\in N$ is a regular value, then the set $f^{-1}(y) \subset M$ is a smooth manifold of dimension $m-n$.
I've some trouble with the very last step.
Proof: Let $x\in f^{-1}(y)$. Since $y$ is a regular value, the derivative $df_x$ must map $TM_x$ onto $TN_y$. The null space $R \subset TM_x$ of $df_x$ will therefore be an $(m-n)$-dimensional vector space.
If $M\subset \mathbb{R}^k$, choose a linear map $L : \mathbb{R}^k \to \mathbb{R}^{m-n}$ that is nonsingular on this subspace $R\subset TM_x \subset \mathbb{R}^k$. Now define
$$F: M \to N\times\mathbb{R}^{m-n}$$
by $F(\xi) = (f(\xi), L(\xi))$. The derivative $dF_x$ is clearly given by the formula $dF_x(v) = (df_x(v), L(v))$. Thus $dF_x$ is nonsingular. Hence $F$ maps some neighborhood $U$ of $x$ diffeomorpically onto a neighborhood $V$ of $(y, L(x))$.Note that $f^{-1}(y)$ corresponds, under $F$, to the hyperplane $y\times \mathbb{R}^{m-n}$.
In fact $F$ maps $f^{-1}(y)\cap U$ diffeomorphically onto $(y\times\mathbb{R}^{m-n})\cap V$. This proves that $f^{-1}(y)$ is a smooth manifold of dimension $m-n$.
If $F$ maps $f^{-1}(y)\cap U$ diffeomorphically onto a an open subset of $\mathbb{R}^p$ it is clear, that $f^{-1}$ is an manifold of dimension $p$. But why $y\times \mathbb{R}^{m-n}$ is of dimension $m-n$? $y$ has dimension $n$ and $\mathbb{R}^{m-n}$ has dimension $m-n$, so the manifold should have dimenson $n$?!
And, a little idea: That $(y\times\mathbb{R}^{m-n})\cap V$ is open subset, because $F$ maps open subsets onto open subsets (because $F$ continuous)?
Best Answer
Note that $y$ is a point in the manifold $N$, although $N$ has dimension $n$, $\{y\}$ has dimension zero, so $\operatorname{dim}\, (\{y\}\times\mathbb{R}^{m-n}) = \dim\, \{y\} + \dim\mathbb{R}^{m-n} = 0 + m - n = m-n$.