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Let $X$ and $Y$ be two locally compact Hausdorff spaces, and let $X^+$ and $Y^+$ denote the one point compactifications of $X$ and $Y$, respectively. Let $f: X\rightarrow Y$ be a continuous function and let $f^+: X^+ \rightarrow Y^+$ be the obvious extension of $f$. Show that $f^+$ is continuous if and only if $f$ is proper.

By proper, I mean that for every compact subset $U$ of $Y$, the preimage of $U$ is compact.

Any help would be appreciated.

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What have you tried? (In either direction.) –  Dylan Moreland Dec 5 '11 at 6:15
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The $+$ notation bothers me, allow me to write $\infty$ for $+$. Suppose first that $f_\infty:X_\infty\to Y_\infty$ is continuous and let $K\subseteq Y$ be compact. We know then that $Y_\infty-K$ is open and so by assumption $f_\infty^{-1}(Y_\infty-K)$ is open. That said, $f_\infty^{-1}(Y_\infty-K)=X_\infty-f_\infty^{-1}(K)$, and by assumption this open. But, note that since $\infty\notin f_\infty^{-1}(K)$ so that $\infty\in X_\infty-f_\infty^{-1}(K)$ and $X_\infty-f_\infty^{-1}(K)$ is open, we have by the definition of the open sets in $X_\infty$ that $f_\infty^{-1}(K)$ is a compact subset of $X$. But, note that $f_\infty^{-1}(K)=f^{-1}(K)$ and thus $f^{-1}(K)$ is a compact subset of $X$.

Conversely, suppose that $f$ is proper and let $U\subseteq Y_\infty$ be an arbitrary open set. If $\infty\notin Y$ then $U$ is open in $Y$ and so $f_\infty^{-1}(U)=f^{-1}(U)$ is open in $X$ and thus in $X_\infty$ by assumption. If $\infty\in U$ then we know that $U=\{\infty\}\cup (X-K)$ for some compact subset of $Y$. Thus, $f_\infty^{-1}(U)=f_\infty^{-1}(\{\infty\}\cup (X-K))=\{\infty\}\cup (X-f^{-1}(K))$. But, since $f$ is proper we know that $f^{-1}(K)$ is compact in $X$ and so $f_\infty^{-1}(U)$ is complement in $X_\infty$ of a compact set in $X$ and thus open. Either way, the preimage of open sets are open, and we're done.

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Any reason why $f_\infty(\infty)\notin Y$ –  Student Dec 5 '11 at 6:49
    
Hmm? What does that mean? Aren't you defining the extension by merely stating $\infty\mapsto \infty$ and everything else is the same? This is how I've always defined it? –  Alex Youcis Dec 5 '11 at 6:54
    
I guess that makes sense. Btw, for the converse, I think you meant $U=\{\infty\}\cup (Y-K)$ Thanks for the answer. –  Student Dec 5 '11 at 6:59
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