Milnor proved that any paracompact Hausdorff space which is equilocally convex (and hence in particular locally contractible) is homotopy equivalent to a CW complex. However, unlike being paracompact and Hausdorff, the property of being equilocally convex seems slightly arbitrary here, while the weaker property of being locally contractible is more conceptual. Does anyone know of a reference for this possible strengthening of Milnor's result, or, possibly, a counterexample?

$\begingroup$ The discussion at mathoverflow.net/questions/102700/… might be useful. $\endgroup$ – Igor Belegradek May 23 '14 at 17:44

$\begingroup$ Also see Martin, John R., Neighborhood contractible spaces. Proc. Amer. Math. Soc. 87 (1983), no. 1, 154–158 $\endgroup$ – Igor Belegradek May 23 '14 at 17:50
After some more digging I found a (somewhat nonexplicit) counterexample to the original question. In his paper "un espace metrique lineaire qui n’est pas un retracte absolu" Cauty constructs a metric vector space $V$ which is not an absolute neighborhood retract. According to the characterization established in "une caracterisation des retractes absolus de voisinage" (also Cauty) a metric space $X$ is an absolute neighborhood retract if and only if each open subset of $X$ is homotopy equivalent to a CW complex. It follows that the metric vector space $V$ above contains an open subset $U \subset V$ which is not homotopy equivalent to a CW. Since $U$ is metrizable it is paracompact and Hausdorff, and since it $V$ is locally contractible so is $U$. Hence $U$ is a counterexample to the original question.
I think Milnor answers your question in the paper where he proves the theorem you refer to ("On spaces having the homotopy type of a CWcomplex", Trans. AMS, 90 (1959), 272280):
Whitehead had observed that any compact space with the homotopy type of a CWcomplex is dominated by a finite complex. It therefore has integral homology in only finitely many degrees. But Borsuk had constructed a locally contractible compact metric space $C$ such that $H_n(C,\mathbb{Z})\neq 0$ for all $n\geq0$.

4$\begingroup$ Borsuk's construction can be found as Theorem 1 here. The idea is to string together spheres of increasing dimension and then compactify that by a Hilbert cube in an appropriate way such that the space remains locally contractible even at the points at infinity. This works because sufficiently small neighborhoods of any point at infinity will not fully contain any of the spheres. $\endgroup$ – Eric Wofsey May 23 '14 at 12:24

3$\begingroup$ Thanks, although there is still a slight problem with this example. It seems that Borsuk is using a somewhat different definition of locally contractible than what I meant. The definition of locally contractible I had in mind was that every neighborhood $U$ of a point $x$ contains a smaller neighborhood $V$ which is itself contractible. Borsuk is using a similar definition but only requiring that the map $V \to U$ will be nullhomotopic. Looking at his example, I cannot tell immediately if it is locally contractible in the stronger sense as well. $\endgroup$ – Yonatan Harpaz May 23 '14 at 14:28

$\begingroup$ Ah, OK. I haven't tried to understand Borsuk's space, but it looks as though @EricWofsey has. Maybe he can say if it's locally contractible in the stronger sense? $\endgroup$ – Jeremy Rickard May 23 '14 at 17:22

2$\begingroup$ I think I managed to convince myself that Borsuk's example is not locally contractible in the strong sense. The space X contains a dense open subset $U$ which is homotopy equivalent to a wedge of spheres. Observing the details of the construction, it seems that a generic point in $X  U$ will have neighborhoods which are equivalent to compactified wedges of spheres, and hence not contractible. $\endgroup$ – Yonatan Harpaz May 23 '14 at 18:24

1$\begingroup$ @YonatanHarpaz: I haven't checked completely, but I think you're right. If I'm not mistaken, a small ball around a typical point in $XU$ will look like a compactification of equatorial strips of some fixed codimension in all of the spheres, and these strips are themselves equivalent to spheres of lower dimension. The only reason this manages to be "locally contractible" in Borsuk's sense is that if you take a slightly larger ball, the codimension of the strips will decrease and so the strips you had before will be contractible inside the new larger strips. $\endgroup$ – Eric Wofsey May 24 '14 at 1:26
It seems that at least a partial answer can be given using the formalism of $\infty$topoi: if $X$ is a paracompact Hausdorff space which is locally contractible (in the strong sense discussed above) and such that the $\infty$topos $Sh(X)$ of sheaves (of $\infty$groupoids) on $X$ is hypercomplete, then $X$ is homotopy equivalent to a CW complex. This happens, for example, when $X$ has finite covering dimension.
To see this, recall Lurie's HTT section 7.1. Given a suitable basis $B$ for the topology on $X$, Lurie constructs a model structure on the category $Top_{/X}$ of spaces over $X$ which is a model for the $\infty$category $Sh(X)$. If $X$ is locally contractible then one can construct a hypercovering of the terminal sheaf on $X$, given concretely by a split simplicial object $U_\bullet$ in $Top_{/X}$ such that each $U_n$ is a coproduct of contractible open subsets of $X$. The geometric realization $U_\bullet \in Top_{/X}$ encodes a sheaf which is $\infty$connective. When $Sh(X)$ is hypercomplete the sheaf $U_\bullet$ admits a section (which in this case will be realized by an actual map $X \to U_\bullet$ in $Top_{/X}$ since in this model structure the terminal object is cofibrant and every object is fibrant) and so $X$ is a retract of $U_\bullet$. Since $U_\bullet$ is a split simplicial object the realization $U_\bullet$ coincides with the corresponding homotopy colimit, and since each $U_n$ is contractible this means that $U_\bullet$ is homotopy equivalent to a CW complex. Since $X$ is a retract of $U_\bullet$ it follows that $X$ is homotopy equivalent to CW complex as well.
This answer is not unsatisfying, but it would be more satisfying if the assumptions above were met by every space which is actually a CW complex. This leads to the following natural question: is $Sh(X)$ hypercomplete for every CW complex $X$?

$\begingroup$ I just posted this as a follow up question: link $\endgroup$ – Yonatan Harpaz May 29 '14 at 14:07