Let $M$ be a manifold and consider an function $f=(f_1,f_2)$: $M\to \mathbb{R}\times\mathbb{R} $. What’s verified is that if either $f_1$ or $f_2$ is a proper map, so is $f$.I’m considering whether $f_1$ and $f_2$ are proper maps when $f$ is proper.
I guess they may not be proper. Let $K \subseteq \mathbb{R}\times\mathbb{R}$ be a compact subset. Then $\pi(K)$ is compact in $\mathbb{R}$ due to the continuity of projection map. We can know that $f^{-1}(K)$ is compact and $f^{-1}(K) \subseteq f_1^{-1}(\pi(K))$. It does not suffice to prove that $f_1^{-1}(\pi(K))$ is compact. I am trying to find some counterexamples but keep failing.
I am looking for an counterexamples and wondering whether they can be proved to be proper when given more condition?
When $f_1$ is proper, then $f$ is proper regardless of whether $f_2$ is proper or not, so $f$ being proper does not imply proper-ness of $f_1$ and $f_2$. Also, consider the identity map $\mathbb{R}^2\to \mathbb{R}^2$, this is a proper map, but the projections are not proper. I am not aware of any conditions that would make $f_1$ or $f_2$ proper given that $f$ is. Although, if $M$ was compact, then any continuous map is proper.