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In Milnor's famous book "Topology from the Differential Viewpoint" he proves the following on page 11:

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 $m-n$.

Right, so I've read this proof previously in another book (using the rank theorem). I cannot however understand one step he does here.

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 $m-n$.


So, I do not understand the part between the *'s. I think I'm sort-of missing something obvious ...

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i think he's saying that $F(f^{-1}(y)) = y \times \mathbb R^{m-n}$. –  Eric O. Korman Jan 28 '11 at 0:10
    
Yes, but how is that evident? That is, how do you show that? –  M.B. Jan 28 '11 at 0:17
    
ahh right, good question! –  Eric O. Korman Jan 28 '11 at 0:28
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yea, it seems like it must be a typo. if $M \subset \mathbb R^k$ is bounded then there's no way that $L(M)$ can be all of $\mathbb R^{m-n}$ so $f^{-1}(y)$ would only correspond to a subset of $y\times \mathbb R^{m-n}$. –  Eric O. Korman Jan 28 '11 at 1:04
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@Eric, @M.B., he is actually saying that $F$ restricted to $f^{-1}(y)$ has image in $y \times \mathbb R^{m-n}$. This is indeed obvious –  Sam Lisi Mar 24 '11 at 11:20
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1 Answer 1

As Eric suggested in the comments, Milnor means simply that $F(f^{-1}(y)) = y \times \mathbb{R}^{m-n}$.

To see this, let $x \in f^{-1}(y) \subseteq M$. Then $f(x) = y$. Also, since $L$ is nonsingular when restricted to the subspace $\mathfrak{N}$ (the nullspace (kernel) of $df_x$, which you wrote as $R$), its image is the entire vector space $\mathbb{R}^{m-n}$.

By definition, $F(x) = (f(x), L(x))$, so it follows that $F(f^{-1}(y)) = y \times \mathbb{R}^{m-n}$.

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