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?

Thanks for your attention

  • $\begingroup$ Just looking at coproducts, the target category would have to be some category of pointed spaces, with coproducts given by wedge sums. $\endgroup$ – Miha Habič Aug 26 '12 at 11:57
  • $\begingroup$ 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. $\endgroup$ – Fosco Aug 26 '12 at 12:02
  • $\begingroup$ I was asking you to check if my argument is correct. Does restricting to proper maps affect colimits in such a way that the one-point compactification functor respects them? I was also asking if the categories I'm considering are "the right ones", in view of the wiki page I quoted... $\endgroup$ – Fosco Aug 26 '12 at 12:10
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    $\begingroup$ 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... $\endgroup$ – t.b. Aug 26 '12 at 12:29
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    $\begingroup$ One should carefully distinguish Alexandroff compactification (that is defined for non-compact spaces only) and Alexandroff extension (that is defined on any topological space, but introduces a new connectivity component if it was already compact). The former is certainly not a functor on LCHaus. $\endgroup$ – Incnis Mrsi Aug 11 '14 at 12:05

Let $\mathsf{LCH}$ be the category of locally compact Hausdorff spaces with proper continuous maps and $\mathsf{CH}$ be the full subcategory of compact Hausdorff spaces with continuous maps. We extend it to the category $\mathsf{CH}_*$ of pointed compact Hausdorff spaces with pointed continuous maps. The Alexandrov compactification is a functor $$A : \mathsf{LCH} \to \mathsf{CH}_*.$$ Of course, we choose $\infty$ to be our base point. Unfortunately, it is not left adjoint to the forgetful functor $U : \mathsf{CH}_* \to \mathsf{LCH}$. However, it has the following universal property:

For $X \in \mathsf{LCH}$ and $(Y,y_0) \in \mathsf{CH}_*$ we say that a continuous map $f : X \to Y$ is proper away from $y_0$ if for all compact subsets $K \subseteq Y$ with $y_0 \notin K$ the preimage $f^{-1}(K) \subseteq X$ is compact. We obtain a set $\mathrm{Hom}_{y_0}(X,Y)$ and it is easy to see that $$\mathrm{Hom}_{\mathsf{CH}_*}(A(X),(Y,y_0)) \cong \mathrm{Hom}_{y_0}(X,Y).$$ Notice that the left hand side (strictly) contains $\mathrm{Hom}_{\mathsf{LCH}}(X,U(Y,y_0))$, the set of proper continuous maps $X \to Y$. It is quite instructive to consider the special case $X=\mathbb{N}$, where maps are just sequences and "properness away from $y_0$" precisely means convergence to $y_0$.

It is easy to check that $A$ preserves finite coproducts. But it does not preserve infinite coproducts. For example, $(A(\{n\}) \to A(\mathbb{N}))_{n \in \mathbb{N}}$ is not a coproduct diagram in $\mathsf{CH}_*$, because this would mean that every sequence converges. In particular, $A$ is not left adjoint to any functor.


Here is one way to get a functor out of Alexandroff compactification, even though it requires significant changes:

Let $\mathsf{LCH}_\hookrightarrow$ be the category of locally compact Hausdorff spaces and open embeddings, and $\mathsf{CH}_*$ the category of pointed compact Hausdorff spaces and pointed continuous maps. Then there is a contravariant functor $$\widehat{(-)} : \mathsf{LCH}_\hookrightarrow^{\rm op} \to \mathsf{CH}_*$$ given by Alexandroff extension$\dagger$ on spaces. If $f : X \hookrightarrow Y$ is an open embedding of LCH spaces, there is an induced map $f^* : \hat{Y} \to \hat{X}$ where: $$f^*(y) = \begin{cases} x & y = f(x), \\ * & y \not\in \operatorname{im}(f). \end{cases}$$ Note that the base point $* \in \hat{Y}$ is not in the image of $f$ so gets mapped to the base point $* \in \hat{X}$. It's not hard to check that this is continuous, because $f$ was an open embedding. Finally, it's clear that the construction is functorial.

$\dagger$ If the space is already compact, this isn't a compactification -- see Incnis Mrsi's comment.

  • $\begingroup$ I realize this doesn't answer the "adjoint" question, but since it's contravariant and not covariant I'm not sure what the adjoint should be... $\endgroup$ – Najib Idrissi Jan 2 '15 at 10:52

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