2
$\begingroup$

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)?

$\endgroup$
  • 1
    $\begingroup$ $y$ (actually, it should be $\{y\}$) is a point. So its dimension is $0$. $\endgroup$ – user228113 Jun 13 '16 at 15:50
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.