I have found 3 links that contain the solution of this question, but in 2 links I have difficulties in understanding some parts of the solution.
For the first link:
The link is given below:
http://www.math.sjsu.edu/~simic/Spring09/Math213/final.213.09.soln.pdf
The theorem proof is given in Q.2 on the third page. but I have a difficulty in understanding the last part of the first line, which says:" then $f^{-1}(y)$" is a discrete subset of $M$"
I could not understand this part, could anyone explain it for me please?
For the second link
The link is given below:
http://www.math.utah.edu/~keenan/manifoldsnotes.pdf
The proof is given in page 35, but I do not understand the following parts of the solution:
1- why he said that $f^{-1}$ is compact, may be he considered that $f^{-1}$ is continuous and he uses the theorem that say that:"A continuous image of a compact set is compact", but then why he considered that $f^{-1}$ is continuous?
2- why he said that $f^{-1}$ is finite?
3- why $f$ is a local diffeomorphism?
Could anyone help me understanding those questions?
I will update the question with the third link and my questions about it later.
Thank you
Best Answer
By the Regular Level Set Theorem, since $y$ is a regular value of $f:X\to Y$, we have that $f^{-1}(y)$ is smooth embedded submanifold of $X$ of codimension (in $X$) equal to the dimension of $Y$. Since $\dim{X}=\dim{Y}$, we have that $f^{-1}(y)$ is a $0$-dimensional submanifold. Since $f$ is continuous, the preimage of any closed set is closed. So $f^{-1}(y)$ is closed, and hence compact, since it's a closed subset of the compact set $X$.
Begin Edit- $f^{-1}(y)$ is discrete since it's a $0$-dimensional manifold. Since it's compact and discrete it must be finite. Indeed, if it was infinite, then it would contain a limit point. Hence every neighborhood of said point would contain other points in the set contradicting discreteness. Hence $f^{-1}(y)=\{x_1,...,x_n\}$. -End Edit
It's a local diffeomorphism about each $x_j\in f^{-1}(y)$ by the inverse function theorem (since $y$ is a regular value $df_{x_j}$ has full rank at each $x_j$).
The remainder of the proof is just chasing open sets until you have one small enough with all the desired properties.