# Thread: Question on something in: Milnor: Topology from the Differentiable Viewpoint

1. ## Question on something in: Milnor: Topology from the Differentiable Viewpoint

Hello,

since I don't want to write down a whole page of the book, you need to have the book to answer me. I'm aware of that and I hope someone has it.

On page 8 Milnor shows that #$\displaystyle f^{-1}(y)$ is locally constant as a function of y.

That means, that for every regular value y, there exists a neighborhood V of y in N, so that every x in V is a regular value (did i get this right?) and #$\displaystyle f^{-1}(y)$=#$\displaystyle f^{-1}(y')$ for every y' in V.

In the last line of the section, he shows a possible neighborhood V of y:
$\displaystyle$V=V_1 \cap ... \cap V_k - f(M-U_1-...-U_k)$$I see that \displaystyle V_1 \cap ... \cap V_k is a neigborhood of y, but why is it still one, if you remove \displaystyle f(M-U_1-...-U_k)$$?

engmaths

2. ## Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

For simplicity we can suppose the neighborhoods U1,...,Uk are all open. So W=M-U1-...-Uk is closed, so W is compact as a closed subset of a compact space M. So f(W) is compact. Again, since N as a manifold is Hausdorff, so f(W) is closed.
So V is open, and it's easy to verify that y belongs to V.

3. ## Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

Thanks, that helped me alot.

Did I get right, that every x in V is a regular value? So that would mean, that for every regular value, there is a whole neighborhood of regular values.

4. ## Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

The set of all critical values are of 0 measure as Sard's theorem says. But I'm not sure about your statement. Sard's theorem says the set of regular values are dense, but it doesn't say it's open.

5. ## Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

I think Milnor already says on page 8 that every x in V is a regular value and therefore every regular value has a neighborhood of regular values. I just ask, because I'm no native speaker and not sure if I got it right. (You don't need to give me a proof.)

edit: V is the neigbhorhood in the starting post.

edit2:
I thought of a proof for: "every x in V is a regular value".
Proof:
$\displaystyle x \in f^{-1}(V)=f^{-1}(U_1 \cap ... \cap U_k - f(M-U_1-...-U_k))$ means $\displaystyle x \in U_i$ for a $\displaystyle i \in \{1,...,k\}$. $\displaystyle f|_{U_i}$ is a Diffeom. and therefore $\displaystyle D(f|_{U_i})(x)$ an Isomorphism. Let i be the inclusion map from U_i to N. Then $\displaystyle f|_{U_i}=f\circ i$ and $\displaystyle D(f|_{U_i})(x)=D(f)(i(x)) \circ D(i)(x)=D(f)(x) \circ D(i)(x)$. So D(f)(x) must be surjective => x is a regular value of f.