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Inspired by the discussion in the comments of this question, I'd like to ask the following question: is it possible to characterize the class of spaces that are homotopy equivalent (or weak equivalent) to compact Hausdorff spaces? As noted in the linked question's comments, no locally connected space with infinitely many components can be homotopy equivalent to a compact Hausdorff space. Are there any other restrictions? Is every path-connected space homotopy equivalent to a compact Hausdorff space? It seems plausible to me that every space might at least be weak equivalent to a compact Hausdorff space: perhaps the topology on an infinite CW-complex can be coarsened to be compact Hausdorff without changing the weak homotopy type.

Update: I've accepted Jeremy Rickard's answer, as it seems to more or less completely answer the case of weak equivalence (amazingly, every space is weak equivalent to a compact Hausdorff space iff there does not exist a measurable cardinal). The comments indicate that spaces having the (strong) homotopy type of compact Hausdorff spaces are much more restricted; I'd still welcome answers elaborating further on this.

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    $\begingroup$ If you allow non-Hausdorff spaces, there are finite path-connected, non-contractible spaces (e.g., the "pseudocircle") which can't be homotopy equivalent to any Hausdorff space, since they have no non-constant continuous maps to any Hausdorff space. This doesn't answer the weak equivalence question, though, as the pseudocircle is weakly equivalent to a circle. $\endgroup$
    – Jeremy Rickard
    Commented Nov 29, 2013 at 15:16
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    $\begingroup$ @JulianRosen: I think so, for weak equivalence at least: take an infinite string of circles glued together like OOOOOOO... and have them shrink and oscillate like a topologist's sine curve that approaches a segment of the first circle. $\endgroup$
    – Eric Wofsey
    Commented Nov 29, 2013 at 16:10
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    $\begingroup$ The paper mimuw.edu.pl/~adamp/preprints/compact/article.pdf proves that iff there are measurable cardinals then there are groups that are not the fundamental group of any compact space. I'm kind of hoping that this is a sledgehammer ... $\endgroup$
    – Jeremy Rickard
    Commented Nov 29, 2013 at 16:39
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    $\begingroup$ @JeremyRickard: Wow, that's fascinating! You should post that as a (partial) answer. From skimming the paper, it seems like what's going on is that if $\kappa$ is measurable, it is possible to define a sort of transfinite composition of length $\kappa$ in the fundamental group of a compact space. $\endgroup$
    – Eric Wofsey
    Commented Nov 29, 2013 at 17:18
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    $\begingroup$ Igor Belegradek has answered the other question. His answer says essentially that a CW-complex which is homotopy equivalent to a compact space must be finitely dominated. In particular, any CW-complex with non-finitely generated homology cannot be homotopy equivalent to a compact topological space. $\endgroup$
    – Ricardo Andrade
    Commented Nov 30, 2013 at 0:41

1 Answer 1

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Expanding on my comment, if there are measurable cardinals then it follows from the results of

A. Przeździecki, Measurable cardinals and fundamental groups of compact spaces. Fund. Math. 192 (2006), no. 1, 87–92.

that there are spaces not weakly equivalent to any compact Hausdorff space, as Przeździecki proves that (if and only if there is a measurable cardinal) there are groups $G$ that are not the fundamental group of any compact Hausdorff space, and so the classifying space of such a group is a counterexample.

He also proves that every group of non-measurable cardinality is the fundamental group of a path-connected compact space, answering a question of Keesling and Rudyak who had earlier proved this with "connected" in place of "path-connected" in

J.E. Keesling and Y.B. Rudyak, On fundamental groups of compact Hausdorff spaces. Proc. Amer. Math. Soc. 135 (2007), no. 8, 2629–2631.

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    $\begingroup$ I'm wondering if Przeździecki's construction of a compact path-connected space with given non-measurable fundamental group actually shows that every non-measurable space is weakly equivalent to a compact space? He starts with a CW-complex with the right fundamental group, and then iteratively takes the Stone-Čech compactification and "path-connectifies". I'm probably missing something, but it seems as though his construction might preserve higher homotopy groups. $\endgroup$
    – Jeremy Rickard
    Commented Dec 1, 2013 at 12:41
  • $\begingroup$ Let me add a few references for the case of countable groups. Any countable group has non-measurable cardinality, so Przeździecki's result you quote applies. Much earlier Felt [Procedings AMS, 1974] showed that any collection of countable groups $\pi_i$ which are abelian for $i>i$ can be realized as homotopy groups of a compact metric space, and in fact, the fundamental group action on higher homotopy groups can be prescribed as well. $\endgroup$
    – Igor Belegradek
    Commented Dec 1, 2013 at 13:27
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    $\begingroup$ @RicardoAndrade: Yes (and in this respect, "non-measurable" means "smaller than the least measurable cardinal", rather than the negation of "measurable" as set theorists usually use the term). When $X$ has measurable cardinality, it is possible that there are unexpected paths in any compactification of $X$ that change the homotopy type, and indeed by the first part of his paper this must happen for spaces with certain fundamental groups. $\endgroup$
    – Eric Wofsey
    Commented Dec 1, 2013 at 21:02
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    $\begingroup$ The discussion above answers everything, but since I received two email invitations... Yes the construction preserves the weak homotopy type of CW complexes of nonmeasurable cardinality. All that is used in the paper is (1) $S^1$ is path connected, even compactness of $S^1$ is not needed (2) If $X$ is a CW complex of nonmeasurable cardinality then $\beta X$ has no nontrivial paths other than those in $X$ (3) A telescope of compact spaces can be compactified without changing its weak homotopy type. -/-/- The construction can be modified so as to include disconnected CW complexes. $\endgroup$
    – Adam Przeździecki
    Commented Dec 2, 2013 at 11:26
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    $\begingroup$ About history - I know only my part in it. I noticed that if any group is a fundamental group of some compact space then Vopenka cardinals can not exist (they are way above measurable cardinals). I thought that `people should know this' and sent a number of queries. Nobody seemed to know; I found a messy argument that it is not the case and dumped it. Keesling and Rudyak turned out to be more interested and published a neat proof below the first measurable cardinal. I took their method and rewrote my part. I should have mentioned the weak homotopy type but I thought nobody would be interested. $\endgroup$
    – Adam Przeździecki
    Commented Dec 2, 2013 at 11:43

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