Prove that on any manifold $M$ there exists a proper smooth map $f: M \to \mathbb R.$

Question

I'm trying to solve the following question:

Prove that on any manifold $M$ there exists a proper smooth map $f: M \to \mathbb R.$

A proper map between topological spaces is a map such that inverse image of compacts are compacts.

There is the following hint: Use $\sigma$-compactness of manifolds and partitions of unity.

I have no idea how to approach this question using partitions of unity. Help?

Answer 1

1. $\sigma$-compactness implies: There exists a locally finite family $(U_n)_{n\in\mathbb N}$ of relatively compact open sets which covers the manifold, i.e. $M = \bigcup_{n\in\mathbb N} U_n$.

2. There exists a smooth partition of unity $(\phi_n)$ subordinate to $(U_n)$.

3. For $x\in M$, set $f(x) = \sum_{n\in\mathbb N} n \cdot \phi_n(x)$. $f$ defines a smooth function on $M$ since $(U_n)$ is locally finite. It is proper, since for any $k\in\mathbb N$ and any $x \notin U_0 \cup \dots \cup U_k$ , \begin{equation} f(x) = \sum_{n>k} n\cdot \phi_n(x) \ge (k+1) \sum_{n>k} \phi_n(x) = k+1, \end{equation} whence $f^{-1}([0,k]) \subset U_0 \cup \dots \cup U_k$ for all $k\in \mathbb N$ – in particular, for any compact $K\subset \mathbb R$, $f^{-1}(K)$ is a closed set contained in some (relatively) compact set of our manifold $M$, and therefore it is compact, too.

The answer should be found also in standard textbooks, such as Bredon or Lee.