Do locally compact Hausdorff spaces and proper maps as morphisms form a category?

Question

Definition: Let $X$ and $Y$ be two topological spaces. A continuous map $f: X \rightarrow Y$ is called proper if $f^{-1}(K)$ is compact for every $K \subset Y$ compact.

Question: If I take the category where objects are locally compact Hausdorff spaces and morphisms are proper maps, does this really form a category?

I think the identity morphism for any object would be just identity proper map, which seems clear. However, I am not sure about the associativity of the composition. For three objects If $f: A \rightarrow B$, $g: B \rightarrow C$ and $h: C \rightarrow D$, where $f, g, h$ are proper maps and $A, B, C, D$ are objects in the category, will $(h \circ g) \circ f = h \circ (g \circ f)$?

Answer 1

Following the Ezio Greggio answer (bel nick fra) you have just to prove that if $f$ and $g$ are proper maps, then $f\circ g$ is proper. This holds of course because $(f\circ g)^{-1}(K)=g^{-1}(f^{-1}(K))$ so if $K$ is compact, then $f^{-1}(K)$ is compact and so $g^{-1}(f^{-1}(K))$ is compact.

Answer 2

The (potential) category $\mathsf C$ of locally compact Hausdorff spaces, with proper maps, is defined as a subcategory of $\mathsf {Top}$; so if $\mathsf C$ is a category, the axioms of associativity and identity hold just because they hold in $\mathsf {Top}$. Basically, you only have to check that $\mathsf {C}$ is closed under the operations of $\operatorname{id}$ and $\circ$; that is, the identities are proper maps, and the composition of proper maps is proper (that is different from the composition of proper maps being associative).