show/hide this revision's text 2 added "finite" twice

Dear Eric, here is a Bourbaki-style proof.

Recall that a continuous map $f: Y\to X$ is called proper by Bourbaki if, for all spaces $Z$, the map $f\times 1_Z: Y \times Z\to X \times Z$ is closed. For example the trivial finite covering $X\times \{ 1,\ldots n \}\to X$ is proper.

Now, your $X$ is covered by opens $X_\iota \subset X$ such that the restricted/corestricted maps $f_{X_\iota }:f^{-1} (X_\iota) \to X_\iota $ are trivial finite coverings, hence are proper by the example above. We deduce that the original covering $f:Y\to X$ is proper: this follows easily from the definition of "proper" and (if a reference is needed) is proved in Bourbaki's General Topology, Chapter 1, §10, Proposition 3.

But a proper map has the property that the inverse image of a quasi-compact subset of the target (in our case all of $X$) is quasi-compact (ibid., Proposition 6). Hence $Y$ is quasi-compact if $X$ is.

NB I have used Bourbaki's definition "universally closed" for proper. As I said, this implies that inverse images of quasi-compact subsets are quasi-compact.This last property is often taken as the definition of proper. For locally compact spaces, both definitions coincide.

show/hide this revision's text 1

Dear Eric, here is a Bourbaki-style proof.

Recall that a continuous map $f: Y\to X$ is called proper by Bourbaki if, for all spaces $Z$, the map $f\times 1_Z: Y \times Z\to X \times Z$ is closed. For example the trivial covering $X\times \{ 1,\ldots n \}\to X$ is proper.

Now, your $X$ is covered by opens $X_\iota \subset X$ such that the restricted/corestricted maps $f_{X_\iota }:f^{-1} (X_\iota) \to X_\iota $ are trivial coverings, hence are proper by the example above. We deduce that the original covering $f:Y\to X$ is proper: this follows easily from the definition of "proper" and (if a reference is needed) is proved in Bourbaki's General Topology, Chapter 1, §10, Proposition 3.

But a proper map has the property that the inverse image of a quasi-compact subset of the target (in our case all of $X$) is quasi-compact (ibid., Proposition 6). Hence $Y$ is quasi-compact if $X$ is.

NB I have used Bourbaki's definition "universally closed" for proper. As I said, this implies that inverse images of quasi-compact subsets are quasi-compact.This last property is often taken as the definition of proper. For locally compact spaces, both definitions coincide.