# Understanding Alexandroff compactification

Is the Alexandroff one-point compactification of a locally compact Hausdorff space ($\mathbf{LCHaus}$) a functor to the category of compact Hausdorff spaces ($\mathbf{CHaus}$)? It seems to me that one has to consider only proper continuous maps as morphisms.

If one does so, then the most natural definition for the induced map $\hat f\colon\hat X\to \hat Y$ seems to work (just send $x\mapsto f(x)$ and $\infty_X\mapsto \infty_Y$, continuity has to be checked only on open sets containing $\infty$) but a lot of interesting situations seem to be excluded: one hopes that (e.g.) a map $h\colon (0,1)\to \mathbb R$ such that $$\lim_{x\to 1^-}h(x)=\lim_{x\to 0^+}h(x)$$ admits an extension $\hat h\colon \widehat{(0,1)}\to \mathbb R$; but what if $h$ is not proper? It is "outside" the category I'm considering. So:

1. How can one define suitable topological categories between which $\widehat{(-)}$ is a functor?

2. Morally a compactification should be a left adjoint to an inclusion (in this case $\iota\colon \mathbf{CHaus}\hookrightarrow\mathbf{LCHaus}$), but even if it seems evident that (taking only proper maps) $$\hom_{\mathbf{CHaus}}(\hat X,Y)\cong \hom_{\mathbf{LCHaus}}(X,\iota Y)$$ the Alexandroff correspondence seems to be ill-behaved with respect to colimits... is there any hope to make $X\mapsto \hat X$ adjoint to something?

Indeed, according to the wikipedia page en.wikipedia.org/wiki/… the Alexandroff extension can be viewed as a functor from the category of topological spaces to the category whose objects are continuous maps $c: X \rightarrow Y$ and for which the morphisms from $c_1: X_1 \rightarrow Y_1$ to $c_2: X_2 \rightarrow Y_2$ are pairs of continuous maps $f_X: X_1 \rightarrow X_2, \ f_Y: Y_1 \rightarrow Y_2$ such that $f_Y \circ c_1 = c_2 \circ f_X$... This seems highly unsatisfactory. –  tetrapharmakon Aug 26 '12 at 12:02
the "evident" adjunction you state can't be true for yet another reason: pre-images of compact sets under proper maps are compact and as $Y$ is compact, $X$ would have to be... –  t.b. Aug 26 '12 at 12:29