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)?

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    $\begingroup$ $y$ (actually, it should be $\{y\}$) is a point. So its dimension is $0$. $\endgroup$ – user228113 Jun 13 '16 at 15:50
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    $\begingroup$ Here $y$ is just a point given at first, not a variable for $f^{-1}$. And you're right in a sense, as $y$ varies, you have the folliated structure of $M$, whose dimension is $n$. $\endgroup$ – cjackal Jun 13 '16 at 15:50
  • $\begingroup$ Another observation: continuous maps do not generally map open sets to open sets (i.e. they may not be open maps): any constant map $f:X\to \Bbb R$ is continuous but not open. What he's using here is the fact that a differentiable map with surjective differential at every point is open (since $dF_x$ is injective and $\dim TM_x=\dim T(N\times \Bbb R^{m-n})_x$, $dF_x$ is surjective). $\endgroup$ – user228113 Jun 13 '16 at 16:01

Note that $y$ is a point in the manifold $N$, although $N$ has dimension $n$, $\{y\}$ has dimension zero, so $\operatorname{dim}\, (\{y\}\times\mathbb{R}^{m-n}) = \dim\, \{y\} + \dim\mathbb{R}^{m-n} = 0 + m - n = m-n$.

  • $\begingroup$ Ok, thats clear. But one question more: why $dF_x$ is nonsingular? $df_x$ and $L$ cannot be nonsingular, because they are not a square matrix. And why it is necessary that $L$ is nonsingular (not a square matrix?) on the kernel? $\endgroup$ – Laura Jun 13 '16 at 16:59
  • $\begingroup$ Recall that we chose $L$ to be non-singular on $R$, the kernel of $df_x$. If $dF_x(v) = 0$, then $df_x(v) = 0$ (so $v \in R$) and $L(v) = 0$. Because $L$ is non-singular on $R$, $L|_R$ is injective, so as $v \in R$ and $L(v) = L|_R(v) = 0$, $v = 0$. Therefore, $dF_x$ is injective, and as $\dim M = \dim N\times\mathbb{R}^{m-n}$, $dF_x$ is also surjective and hence an isomorphism. In particular, $dF_x$ is non-singular. $\endgroup$ – Michael Albanese Jun 13 '16 at 18:04
  • $\begingroup$ But $dF_x$ is only on $R$ non-singular isn't it? Because otherwise L is not non-singular. $\endgroup$ – Laura Jun 13 '16 at 18:30
  • $\begingroup$ $L$ may not be non-singular (neither Milnor nor I claim it is non-singular). However, we only need $L$ to be non-singular on $R$. $\endgroup$ – Michael Albanese Jun 13 '16 at 18:32

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