Inverse image of compact is compact Let $f : X\to Y$ be a closed map of topological spaces, such that the inverse image of each point in $Y$ is a compact subset of $X$. Is it true that the pre-image of a compact set $K$ is compact?
The answer is yes but I’m not sure how to show it. I just know that the inverse image of a singleton, say $y,$ is contained in the union of finitely many sets which we can call $U_y$. That’s it. Why is the statement of the question true?
I saw this Why is the inverse image of a compact set under a special sort of function compact?
But it’s too old to revive. Can someone post an easy to follow solution. I’ve just read about the finite intersection property but to understand it I had to come read it here:
http://mathonline.wikidot.com/finite-intersection-property-criterion-for-compactness-in-a
 A: Suppose $C \subseteq Y$ is compact.
Let $\mathcal{U}$ be an open cover for $f^{-1}[C]$ by open sets of $X$, so that
$$f^{-1}[C] \subseteq \bigcup \mathcal{U}$$
For each $y \in C$ we know that $f^{-1}[\{y\}]$ is compact by assumption and as $f^{-1}[\{y\}]\subseteq f^{-1}[C]$, this compact set is also covered by $\mathcal{U}$ and so for each $y \in C$ we have a finite subset $\mathcal{U}_y$ of $\mathcal{U}$ such that:
$$f^{-1}[\{y\}] \subseteq U_y := \bigcup \mathcal{U}_y \tag{1}$$
Next we define $$O_y = Y\setminus f[X\setminus U_y]\tag{2}$$
and note that $$\forall y \in C: y \in O_y\tag{3}$$
and $$\forall y \in C: f^{-1}[O_y] \subseteq U_y\tag{4}$$
Proof of (3): suppose that $y \notin O_y$. This means by def. $(2)$ that $y \in f[X\setminus U_y]$, so there is some $x \in X\setminus U_y$ with $f(x)=y$. But then this $x \in f^{-1}[\{y\}]$ but $x \notin U_y$, contradicting $(1)$. This shows $(3)$.
Proof of (4): let $x \in f^{-1}[O_y]$ be arbitary. If $x \notin U_y$ we'd have that $f(x) \in f[X\setminus U_y]$ so $f(x) \notin O_y$ contradicting that $x \in f^{-1}[O_y]$. So $x \in U_y$ and the inclusion $(4)$ has been shown.
So $\{O_y: y \in C\}$ is an open cover of $C$ and as $C$ is compact, there are finitely many $y_1, y_2, \ldots y_n \in C$ such that $C \subseteq \bigcup_{i=1}^n O_y$
Then (using $(4)$ as well): $$f^{-1}[C] \subseteq f^{-1}[\bigcup_{i=1}^n O_y] = \bigcup_{i=1}^n f^{-1}[O_y] \subseteq \bigcup_{i=1}^n \bigcup \mathcal{U}_{y_i}\tag{5}$$
and $(5)$ shows that $\bigcup_{i=1}^n \mathcal{U}_{y_i}$ is a finite (a finite union of finite sets) subcover of $\mathcal{U}$, as we had to find. So $f^{-1}[C]$ is compact.
Note that continuity of $f$ is never used, just ontoness and closedness of $f$ and its compact fibres..
