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In a algebraic topology course, I recently saw the proof of the theorem that the $n-$ and $m-$ dimensional vector spaces, $n\neq m$, are not isomorphic (in $\mathbf{Top}$). The proof used the compactification of both of these spaces (which are $n-$ and $m-$spheres respectively) and then the reduced homology functor which shows that these are not isomorphic. For this to work it is necessary that the compactification preserves isomorphisms. It was shortly (in elementary terms) explained why that is the case. I wonder now wether the compactification is functorial in general, or wether it just preserves isomorphisms in this specific case.

tl;dr: Question is the title.

Guenterino
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  • What is true and is easy to verify: if $X$ is homeomorphic to $Y$ then $\alpha X $ (the Aleksandrov, or one-point compactification of $X$) is homeomorphic to $\alpha Y$. – Henno Brandsma Oct 16 '21 at 08:51
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    It's no true that a continuous $f: X \to Y$ always gives rise to a map $\alpha f: \alpha X \to \alpha Y$ e.g. So it's not functorial in the proper sense. It's just a topologically unique compactification. – Henno Brandsma Oct 16 '21 at 08:53
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    Do you consider the one-point compactification of locally compact Hausdorff spaces or the Alexandroff one-point compactification for arbitrary spaces? – Paul Frost Oct 16 '21 at 08:58
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    The Alexandroff compactification is a functor from the category $\mathsf{LCH}$ of locally compact Hausdorff spaces and proper continuous maps, to the subcategory $\mathsf{CH}_*$ of pointed compact Hausdorffs, see this question. – user10354138 Oct 16 '21 at 09:05
  • @PaulFrost I didn't make a distinction so far. If it is functorial for some category but not for all that would answer my question - which is what you did in your answer, so thank you for that! :) – Guenterino Oct 17 '21 at 09:09

2 Answers2

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As Henno Brandsma said in a comment, it is not functorial.

The Alexandroff one-point compactification of an arbitrary space $X$ is defined as the set $X^* = X \cup \{\infty\}$ with a point $\infty \notin X$. The topology on $X^* $ consists of all open subsets of $X$ and all sets of the form $X^* \setminus C$, where $C \subset X$ is closed and compact. The sets $X^* \setminus C$ are precisely the open neigborhoods of $\infty$.

Let us consider a map $f : X \to Y$. When does it extend to a continuous $f^* : X^* \to Y^*$ such that $f^*(\infty) = \infty$?

Clearly $f^*$ is continuous iff for each closed compact $D \subset Y$ we have $(f^*)^{-1}(Y^* \setminus D) = X^* \setminus C$ for some closed compact $C\subset X$. But $(f^*)^{-1}(Y^* \setminus D) = X^* \setminus f^{-1}(D)$, thus $f^*$ is continuous iff preimages of closed compact $D \subset Y$ under $f$ are compact (they are automatically closed by continuity). Such maps $f$ are called proper. Therefore the one-point compactification is a functor from the category of topological spaces and proper maps to the category of compact spaces and continuous maps.

Note that all homeomorphisms are proper. In contrast, constant maps are never proper unless the domain $X$ is compact.

Paul Frost
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It is not functorial. But the one-point compactification of two homeomorhpic spaces is still homeomorphic.

A simple counterexample to functoriality is the inclusion map $\mathbb Q\to\mathbb R.$ The open subsets around $\infty$ in $\mathbb Q^*$ are all dense in $\mathbb Q^*,$ but that isn't true if the image of $\mathbb Q^*$ in $\mathbb R^*.$

Basically, subspaces can have very different classes of compact subsets. $\mathbb Q$ has much fewer compact subsets than $\mathbb R.$

Thomas Andrews
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  • In what sense do you mean that $\mathbb{Q}$ has fewer compact subsets than $\mathbb{R}$? – Jakobian May 28 '25 at 00:44
  • Given any compact subset of $\mathbb Q,$ it is compact in $\mathbb R,$ but there are many compact $K\subseteq\mathbb R$ it is not true that $K\cap\mathbb Q$ is compact in $\mathbb Q.$ @Jakobian – Thomas Andrews May 28 '25 at 01:59
  • Oh okay. Because I thought that maybe you mean cardinality, but both spaces have the same amount of compact sets in this sense. So I suppose you meant it on an intuitive level, which I have the same intuition as you about it. – Jakobian May 28 '25 at 02:03
  • Yeah, it was meant to be intuitive. @Jakobian – Thomas Andrews May 28 '25 at 02:15