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The following question is taken from Basic Topology by Armstrong:

Let $f$ be a closed map with the property that the inverse image of each point of $Y$ is a compact subset of $X$. Show that $f^{-1}(K)$ is compact whenever $K$ is compact in $Y$.

I've written my attempt at proving this problem below, but I'm not sure if it's completely correct (especially concerned about correctness near the end of the proof). Any comments/ideas for a nicer proof would be appreciated!

Proof. Let $\mathcal{U}$ be an open cover of $f^{-1}(K)$. Since $f^{-1}(\{y\})$ is compact for any $y \in Y$, we can reduce the cover above to $\{U_y\}$, where $U_y$ is a finite union of open sets in $\mathcal{U}$ such that $f^{-1}(\{y\}) \subseteq U_y$. Observe that $y \in f(U_y^C)^C$, and so $\{f(U_y^C)^C \cap K\}$ is an open cover of $K$. This reduces to a finite subcover by compactness of $K$, so $K \subseteq \bigcup_{i=1}^{n}f(U_{y_i}^C)^C$ for some $y_1, \dots, y_n \in Y$. Finally, observe that $f^{-1}(f(U_{y_i}^C)^C) = U_{y_i}$, and so $\{U_{y_i}\}$ is an open cover of $K$. By construction of $U_{y_i}$, $f^{-1}(K)$ is compact by definition. $\square$

Kurtland Chua
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    It's not true that if $x \in U$ that necessarily $f(x) \in f(U^c)^c$. For example, consider the unique map ${a,b} \to {x}$. Then if $U= {a}$ we have $f(U^c)=f(U)$. It is true however that $f^{-1}(f(U^c)^c) \subseteq U$, since if $f(x) \in f(U^c)^c$, either $x \in U$ or $x \in U^c$. But if $x \in U^c$ we have $f(x) \in f(U^c)$, a contradiction. This still allows you to conclude as you do – Exit path Feb 21 '18 at 01:40
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    @leibnewtz Doesn't it work in this particular case? $U_y$ contains the preimage of $y$ by construction, so that counterexample wouldn't work? – Kurtland Chua Feb 21 '18 at 01:57
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    Well mostly the point is that you'd have to make some sort of extra argument to show equality, which as I pointed out isn't necessary – Exit path Feb 21 '18 at 02:25
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    You're not using that $f$ is closed. – Henno Brandsma Feb 21 '18 at 05:19
  • @HennoBrandsma I used it in saying that $f(U^C)^C$ is open? – Kurtland Chua Feb 21 '18 at 05:20
  • Indeed, it's implicitly using the lemma I mentioned. – Henno Brandsma Feb 21 '18 at 19:37

1 Answers1

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Useful lemma for a correct proof:

A surjective $f: X \to Y$ is a closed map, iff for every $y \in Y$ and every $O$ open in $X$ such that $f^{-1}[\{y\}] \subseteq O$ there is some $U$ open in $Y$ that contains $y$ and such that $f^{-1}[U] \subseteq O$.

For the forward direction (which we need) one indeed follows the OPs' idea of taking $U = Y\setminus f[X\setminus O]$ and shows it fits the bill.

Then the OP's idea works but can be written in a clearer notation:

For compact $K$ and an open cover $\mathcal{U}$ of $f^{-1}[K]$ we cover each $y \in K$ the compact $f^{-1}[\{y\}]$ by a union of finitely many members of the cover $\mathcal{U}$ and call this union $O_y = \cup \mathcal{U}_y$, for a finite $\mathcal{U}_y \subseteq \mathcal{U}$. Then apply the lemma in forward direction and we get $U_y \ni y$ open in $Y$ with $f^{-1}[U_y] \subseteq O_y$. This gives an open cover of $K$ by the $U_y, y \in K$ so finitely many of these cover $K$, say $U_{y_1},\ldots, U_{y_n}$, and then $\cup_{i=1}^n \mathcal{U}_{y_i}$ is a finite subcover of $\mathcal{U}$ for $f^{-1}[K]$.

Henno Brandsma
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  • Didn't I essentially use this fact in the proof above? In my case, $O = U_y$, and $U = f(U_y^C)^C$, where $U$ is open since $U_y^C$ is closed, and is thus mapped to a closed set by the closedness of $f$. – Kurtland Chua Feb 21 '18 at 05:30