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On page 11 of Milnor's Differential Topology book there is Lemma 1.

In the proof of Lemma 1 it says, to define, $ F:M\to N\times \mathbb{R}^{m-n}$ by $F(\xi) = (f(\xi),L(\xi))$. The derivative produces $dF_x(v) = (df_x(v),L(v))$, so far this is all good. The part I do not see is why does $f^{-1}(y)$ correspond to $y\times \mathbb{R}^{m-n}$. I see why the first coordinate $y$ is there since the first coordinate of $F$ is $f$ which will send $f^{-1}(y)$ to $y$, but how do we know that $L$ sends $f^{-1}(y)$ to all of $\mathbb{R}^{m-n}$?

Nicolas Bourbaki
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1 Answers1

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$L$ is defined to be non-singular on a subspace of dimension $n-m$. Hence it is an isomorphism, thus onto. It's linear. So, it's its own derivative.

EDIT: Milnor is not saying that $F$ maps $f^{-1}(y)$ onto $y\times\mathbb{R}^{n-m}$. He only means that it maps into. Check out the circle, defined by $f^{-1}(1)$ for $$ f(x,y)=x^2+y^2 $$ Then $(1,0)\in f^{-1}(1)$. You can then check that you may define $L(v_1,v_2)=v_2$. Then $F:S^1\rightarrow 1\times\mathbb{R}$, $F(x,y)=(1,y)$, is certainly not onto, because $-1\leq y\leq 1$ when $(x,y)\in S^1$.

J126
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  • $L$ is non-singular on this subspace. But we are saying that $L$ sends $f^{-1}(y)$ onto $\mathbb{R}^{m-n}$, this is where my problem is. Clearly $L$ sends its subspace to $\mathbb{R}^{m-n}$ because it is non-singular but why does is still send $f^{-1}(y)$ to $\mathbb{R}^{m-n}$? – Nicolas Bourbaki Feb 22 '14 at 20:44