# Proper map and manifolds

Let $M$ and $N$ two manifolds which have the same dimension, $f:M\to N$ a map $\mathcal{C^\infty}$. We suppose that $M$ is compact and we have $b$ a regular value of $f$.

First, I have to prove that $f^{-1}(\{b\})$ is a compact and finite set.

For compacity, I thought about the topology of $\mathbb{R}^m$ using the fact that if a set is closed and bounded it's compact. I tried to define a map $\mathbb{R}^m\to \mathbb{R}^m$ using homeomorphism $\phi$ and $\psi$ but I don't know if it is a good method. To prove that it is finite I have no idea.

Second, I have to prove that there exists an open set $V$ of $N$ which contains $b$ such as for all $c\in V$ we have $\mid f^{-1}(\{c\}) \mid\ge\mid f^{-1}(\{b\}) \mid$.

Thanks in advance !

• If $b$ is a regular value, $f^{-1}(b)$ is a manifold of dimension $\text{dim} M - \text{dim} N = 0$ by the local submersion theorem. So that's a discrete set of points. That's a closed set because it's a level set. Now $M$ is compact, so it's compact as it's a closed subspace. Compact discrete sets are always finite. – Balarka Sen Jun 3 '16 at 18:45

## 1 Answer

Since $f$ is continuous, $f^{-1}(b)$ is closed. It is compact since it is a closed subset of a compact set.

Let $x\in f^{-1}(b)$, since $f$ is regular, $df_x:T_xM\rightarrow T_bN$ is an isomorphism. You can work locally and restrict $f$ to a neighborhood $U$ of $x$ such that $f(U)\subset V$ where $V$ is a neighborhood of $b$ and $U$ and $V$ are diffeomorphic to open subsets of $R^n$. This implies that the differential $f_{\mid U}:U\rightarrow V$ is invertible so it is a local diffeomorphism. We deduce that there exists an open subset $x\in U_x$ such that the restriction of $f$ to $U_x$ is a diffeomorphic onto its image. This implies that $f^{-1}(b)\cap U_x=\{x\}$. Thus $f^{-1}(b)$ is discrete. A discrete compact set is finite.

• I don't understand the last implication – Maman Jun 3 '16 at 15:59
• If $X$ is discrete and compact, suppose that $X$ is infinite, you can construct a sequence $(x_n)$ of points pairwise distinct. It has an accumulation point $x\in X$, but you have $U_x$ such that $U_x$ is open and $U_x=\{x\}$, thus since there exists $N$ such that $n>N$ implies $x_n\in U_x$, $x_n=x, n>N$ this is a contradiction with the fact that the points of the sequence are pairwise distinct. – Tsemo Aristide Jun 3 '16 at 16:07
• For the second question, we can say that if $c$ is not a regular value of $f$ then $f^{-1}(\{c\})$ is not finite ? – Maman Jun 3 '16 at 17:14
• It is possible, take for example the constant map whose image is $y$, $y$ is not a regular value and it's inverse image maybe infinite – Tsemo Aristide Jun 3 '16 at 17:17
• And for the open set I would choose $f_{\mid U} (U)=V$, the equality is given if we take $c=b$ and for the other cases $f^{-1}(\{c\})$ which is infinite. But the inverse image for $c$ is not necessary in $U$. – Maman Jun 3 '16 at 17:21