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How can I prove that if $f:X \to Y$ is continuous of locally compact, Hausdorff topological spaces, then $f$ is proper (inverses of compact sets are compact) iff it extends continuously as a map between the one point compactifications of $X$ and $Y$.

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  • $\begingroup$ You need $X$ and $Y$ to be locally compact Hausdorff in order for them to have one-point compactifications. $\endgroup$ – Qiaochu Yuan Sep 15 '14 at 1:40
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I don't provide a full solution, but leave it to you to do the last step.

Assume that $f$ is proper. Then we define $\hat f:\hat X\to\hat Y$ by $\hat f|_X=f$ and $\hat f(\infty_X)=\infty_Y$. Since $f$ is continuous on $X$ and $X$ is open in $\hat X$, the extension $\hat f$ is continuous at each point in $X$. So we only need to check the continuity at $\infty_X$. Take any open neighborhood $V$ around $\infty_Y$. What do you know about its complement and what can you conclude about its preimage?

Conversely, assume $f$ extends to a map $\hat f: \hat X \to\hat Y$ by setting $\hat f(\infty_X)=\infty_Y$. If $K$ is a compact set in $Y$, what does this mean for $\hat Y\setminus K$, and what does the continuity of $\hat f$ imply then?

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  • $\begingroup$ For the first part...The complement is compact, correct? So it's preimage is compact, so the complement of it's preimage is open. For the second part...the complement of $K$ is open, so the continuity of our new function implies the preimage is open, but then the complement of the preimage is compact. Am I correct so far? $\endgroup$ – Johnny Apple Sep 16 '14 at 19:17
  • $\begingroup$ Yes, and use in both parts that the complement of the preimage is just the preimage of the complement @JohnnyApple. $\endgroup$ – Stefan Hamcke Sep 16 '14 at 19:26

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