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Let $X,Y$ be Hausdorff and locally compact topological spaces, with $Y$ not compact, and let $\hat{Y}$ be the Alexandroff compactification of $Y$. Let $A\subset X$ be an open, relatively compact subset of $X$ and consider a map $f:A\to Y$. Denote by $\hat f:X\to \hat Y$ the extension of $f$ defined as follows:

\begin{equation*} \hat f(X-A)=\infty \end{equation*} (while, if $x\in A$, $\hat f(x)=f(x)$).

Show that $\hat f$ is continuous if and only if $f$ is continuous and proper, that is:

\begin{equation*} K\subset Y\text{ compact }\Longrightarrow f^{-1}(K)\subset A \text{ compact} \end{equation*}

I have proved that if $\hat f$ is continuous, then $f$ is continuous too. However, I am having some troubles showing that $f$ is also proper: if $K\subset Y$ is compact, then, since $Y$ is Hausdorff, $K$ is closed in $Y$.. Since $f$ is continuous, $f^{-1}(K)\subset A$ is closed in $A$. Now, if $f^{-1}(A)$ were closed also in $\overline{A}$, I could conclude that $f^{-1}(K)$ is compact, because it would be closed in a compact set. However, since $A$ is an open subset of $X$, then $A$ is open in its closure, because: \begin{equation*} A=\overline{A}\cap A \end{equation*} So, $f^{-1}(K)$ closed in $A$ does not necessarily implies that $f^{-1}(K)$ is closed in $\overline{A}$ too (the implication would be true if $A$ were closed in its closure).

Do you have any suggestion?

Thanks for any help!

Steppenwolf
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  • If $K \subset Y$ is compact, you can separate it from $\infty$ by open sets in $\hat{Y}$ (that is Hausdorff since $Y$ is locally compact). So if $\hat{f}$ is continuous, look at the preimages of the separating open sets. – Dermot Craddock Oct 26 '25 at 11:12
  • I need $\hat Y$ to be regular in order to separate a closed set $K$ from a single point (that is, the infinity point), but I do not think $\hat Y$ is regular, in general. – Steppenwolf Oct 26 '25 at 11:19
  • Since $\hat{Y}$ is compact (quasicompact and Hausdorff), it is regular (completely regular, and normal). In general, though, since $K$ is not only close but compact, the Hausdorff property is enough. In Hausdorff spaces, you can separate compact sets from points (outside the compact set) with open sets. – Dermot Craddock Oct 26 '25 at 12:30
  • Okk, I am sorry. I had forgotten this property of the Hausdorff and compact spaces....however, I cannot see how this property can help me showing the preimage of $K$ is closed in the closure of $A$ – Steppenwolf Oct 26 '25 at 13:57

1 Answers1

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Ok, I've got it:

let $K\subset Y$ be a compact subset of $Y$. Since $Y$ is Hausdorff, $K$ is also closed in $Y$; since $f$ is continuous, $f^{-1}(K)\subset A$ is closed in $A$. On the other hand, by definition of Alexandroff compactification, $K$ is also closed in $\hat Y$. Since $\hat f$ is continuous, then $\hat f^{-1}(K)=f^{-1}(K)\subset X$ is a closed subset of $X$. Then, since $f^{-1}(K)\subset A$ and $f^{-1}(K)$ is closed in $X$, \begin{equation*} f^{-1}(K)\subset \overline{A} \end{equation*} where $\overline{A}$ is the closure of $A$ in $X$. Thus, \begin{equation*} f^{-1}(K)=\overline{A}\cap f^{-1}(K) \end{equation*} is closed in $\overline{A}$. Since, by hypotheses, $\overline{A}$ is compact and $f^{-1}(K)$ is closed in $\overline{A}$, then $f^{-1}(K)$ is compact. This means that $f$ is proper as desired.

Now, there is another problem: for the converse, assume $f:A\to Y$ is continuous and proper and let $B\subset \hat Y$ be open. If $B$ is an open subset of $Y$, then everything is fine, but if $B=\hat Y-K$ for a certain $K\subset Y$ compact subset of $Y$, then: \begin{equation*} \hat f^{-1}(\hat Y-K)=X-\hat f^{-1}(K)=X-f^{-1}(K) \end{equation*} Now, since $Y$ is Hausdorff, $K$ is also closed in $Y$, so $f^{-1}(K)$ is closed in $A$, but it is not necessarily closed in $X$. Thus, we cannot conclude that $\hat f^{-1}(B)$ is open in $X$...

I can also say that, since $f$ is proper, then $f^{-1}(K)$ is compact, but the compactness is referred to the topology induced by $A$, not by $X$ so, even if $X$ is Hausdorff, I cannot conclude that $f^{-1}(K)$ is closed in $X$, right?

Okk, I've got it: if $f^{-1}(K)$ is compact with the topology induced by $A$, actually it is compact also with the topology induced by $X$, so: yes, $\hat f^{B}$ is open in $X$, thus $\hat f$ is continuous and this concludes the proof.

Steppenwolf
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