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Does a compact semilocally simply connected geodesic space have the homotopy type of a compact CW complex? Actually what I'd like to know is whether the fundamental group of such a space is finitely presented.

Edit: As Bruno Martelli notes, this is obviously false, but the question of whether the fundamental group is finitely presented, which is what I really want to know, is still open.

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  • $\begingroup$ I don't understand the comment in your edit: The fundamental group of the Hawaiian earrings is known to be uncountable, hence it cannot be finitely presented, at least as a discrete group. Could you clarify what you mean? $\endgroup$
    – Theo Buehler
    Commented Jan 12, 2011 at 16:25
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    $\begingroup$ The Hawaiian earrings is not semi-locally simply-connected. $\endgroup$
    – Bruno Martelli
    Commented Jan 12, 2011 at 17:38
  • $\begingroup$ Do semilocally simply connected, geodesic, and compact imply that small balls are simply connected? The "counterexamples" I can think of aren't geodesic. If you have a locally finite cover by absolute retracts (ARs) with all intersections ARs, then Weil's theorem says the nerve is homotopy equivalent to the space. So maybe you could do a similar thing with a covering by simply-connected sets to get a compact nerve whose fundamental group surjects that of the space. (Local simple-connectivity would allow a map from the 2-skeleton of the nerve to the space.) $\endgroup$
    – Autumn Kent
    Commented Jan 12, 2011 at 18:44
  • $\begingroup$ The cone over the Hawaiian earrings is semi-locally simply-connected and also geodesic, right? This space is not locally simply-connected. $\endgroup$
    – Bruno Martelli
    Commented Jan 12, 2011 at 18:56
  • $\begingroup$ @Bruno, I think the cone over the Hawaiian earring is not geodesic, because there are points that are very close together (at height 1/2 in the cone) that are not joined by a short geodesic. (I could be wrong, I'm easily confused thinking about weird spaces.) $\endgroup$
    – Autumn Kent
    Commented Jan 12, 2011 at 19:01

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The bouquet of infinitely many shrinking 2-spheres in $\mathbb R^3$ centered in $(0,0,n)$ and of radius $n$ is a compact simply connected geodesic space, which is not homotopic to a compact CW complex.

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I hope I have not goofed, but I think the answer to your modified question is yes:

The fundamental group of a semi-locally simply connected, compact and geodesic space is finitely presented.

Here are the ingredients - all numbers in parentheses refer to Bridson-Haefliger, Metric spaces of non-positive curvature, Springer Grundlehren, 1999, Part I.

  • The universal covering space $\widetilde{X}$ equipped with the length metric induced by the covering projection is a length space (3.25).
  • The fundamental group $\pi_{1}(X)$ acts on $\widetilde{X}$ properly and cocompactly by isometries (8.3 (2)), see also (8.5).
  • If a length space $\widetilde{X}$ admits a proper and cocompact action by isometries then it is locally compact (8.4 (1)) and hence proper and geodesic (3.7).
  • A group is finitely presented if and only if it acts properly and cocompactly by isometries on a simply connected geodesic space (8.11).

All this taken together yields that $\widetilde{X}$ is a simply connected geodesic metric space and $\pi_{1}{(X)}$ acts properly and cocompactly by isometries, hence $\pi_{1}{(X)}$ must be finitely presented.

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  • $\begingroup$ I'd mark this answer correct too if I could! $\endgroup$
    – Jim Conant
    Commented Jan 12, 2011 at 19:52
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In Bourbaki's Topologie Algébrique text, a space is called délaçable* if it is locally path-connected and semi-locally simply-connected.

Theorem 1 in chapter IV, §2, No.3 states that

Let $X$ be a compact délaçable space and $x \in X$. Then $\pi_1(X,x)$ is finitely presented.

*I have seen unloopable for its English but that does not seem like a common use.

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