I've been trying to prove the following result to no avail:
Let $f: X \to Y$ be a continuous, closed, surjective map with compact fibres (i.e. a perfect map) between two topological spaces $X$ and $Y$. Moreover, suppose that $Y$ is Lindelöf. Prove that $X$ is Lindelöf.
So far, I tried pushing an open cover of $X$ into Y, constructing a countable subcover and pulling that back into $X$, but I was not able to show that this would indeed cover X. I also tried constructing a countable compact cover of the $f^{-1}(\{y\})$. These approaches I also attempted using the map $f^{\ast}(U) = Y/f(X/U)$ and its various nice properties.
A hint would be much appreciated!
Thank you
Best Answer
Let $\mathcal{U}$ be an open cover of $X$. WLOG we can assume it's closed under finite unions (if we extend the cover by these finite unions, if the larger cover has a countable subcover, so has the original one).
It then follows, by closedness of the map $f$ and compactness of its fibres (or also from your $f^\ast$) that for every $y \in Y$ we can find an open $y \in V_y \subseteq Y$ so that $f^{-1}[V_y] \subseteq U_y$ for some $U_y \in \mathcal{U}$.
Now the $U_y$, corresponding to the countable subcover of the $V_y$ of $Y$, are as needed.