My question is about a property of proper maps.
We say $f : X \to Y$ es proper if for $C \subset Y$ compact is $f^{-1}(C) \subset X$ compact.
Given $f_1 : X \to Y_1, f_2 : X \to Y_2$ differential maps between differential manifolds, prove that if $f_1$ is proper, then $f:=(f_1,f_2) : X \to Y_1 \times Y_2$ is proper.
I get:
Let be $C_1 \times C_2 \subset Y_1 \times Y_2$ compact. Since $Y_1 \times Y_2$ is Hausdorff we get $C_1 \times C_2$ es closed. Since $f$ is continious ($f_1,f_2$ are differentiable)$ f^{-1}(C_1 \times C_2)$ is closed in $M$.
Also, since $f_1$ is compact, $f^{-1}(C_1) \subset M$ is compact.
And how the argument it follows?
Thanks!
Denote by $pr_j$ the projection from $Y_1 \times Y_2$ to $Y_j$ for $j\in \{1,2\}$. Then we have $$ f^{-1}(C) \subseteq f_1^{-1}(pr_1(C)) \cap f_2^{-1}(pr_2(C)). $$ Now use that $pr_j$ are continuous and that compact sets get mapped to compact sets under continuous maps. Thus, the RHS is compact. But the LHS is closed (as it is the preimage of a closed set under a continuous map) and hence compact as well.