Continuity of the extension of a function between two locally compact Hausdorff spaces to their one point compactifications 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.
 A: 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 $U=\{\infty\}\cup (Y-K)$ for some compact subset $K$ 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.
