Let X be a Hausdorff space such that every real vector bundle on X is summand of a trivial bundle. Does this imply that X is homotopy equivalent to a compact Hausdorf space? This question is a "compact version" of the following question;
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asked Nov 28, 2013 at 14:47
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1$\begingroup$ If $X$ is a smooth manifold, then every vector bundle over $X$ is a summand of a trivial bundle, i.e. embed the total space of the vector bundle in some $\mathbb R^N$, and restrict the normal bundle to the zero section. Is every manifold homotopy to a compact Hausdorff space? This seems unlikely. $\endgroup$– Igor BelegradekCommented Nov 28, 2013 at 20:34
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3$\begingroup$ I don't know any restrictions on the homotopy type of a compact Hausdorff space. Can anyone name one? $\endgroup$– Qiaochu YuanCommented Nov 28, 2013 at 20:53
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2$\begingroup$ An infinite discrete space is a counterexample, but finding a connected counterexample seems harder: I don't know how to show a connected space does not have the homotopy type of any compact Hausdorff space. $\endgroup$– Eric WofseyCommented Nov 28, 2013 at 22:04
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3$\begingroup$ @Qiaochu Yuan: A space homotopy equivalent to a discrete space (or a CW-complex, or any space with open path components) has open path components. Thus any space homotopy equivalent to an infinite discrete space would have an infinite open cover by pairwise disjoint, non-empty subsets. Such a cover admits no finite subcover. $\endgroup$– Ricardo AndradeCommented Nov 28, 2013 at 23:05
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1$\begingroup$ @Ricardo: oh, I see. If $X$ is such a space and $f : X \to D$ a homotopy equivalence to a discrete space then in particular $f$ is a continuous map inducing an isomorphism on $\pi_0$, so the inverse image of each point in $D$ is open and these must be the path components of $X$. Thanks! (If we weaken to weak homotopy equivalence then this argument fails, though, and in this case maybe everything has the weak homotopy type of a compact Hausdorff space? It's just that I'm hesitant to talk about the homotopy type of something that isn't homotopy equivalent to a CW complex.) $\endgroup$– Qiaochu YuanCommented Nov 28, 2013 at 23:17
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1 Answer
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Let $X$ be the wedge of infinitely many circles (equipped with the CW topology). Every vector bundle $\xi$ over $X$ is a summand of a trivial bundle, namely it is $\xi\oplus\xi$ is trivial because any vector bundle over a circle has this property (alternatively, one could appeal to the fact that $X$ is homotopy equivalent to a smooth manifold, if the number of circles is countable, as explained in my comment above).
Suppose there is a compact space $K$ and a homotopy equivalence $f: K\to X$. The image $f(K)$ is compact, so $f(K)$ lies in a finite subcomplex $X_0$ of $X$, i.e. $X_0$ is a wedge of finitely many circles. Thus any loop on $X$ is freely homotopic to a loop in $X_0$, which is false (because a circle that forms the wedge can be homotoped into $X_0$ only if it lies in $X_0$.
EDIT: A key feature of the above example is that there is a homotopic to the identity $X\to X$ whose image lies in a compact subset, and it allows for the following optimal generalization.
Let $X$ be a (say path-connected) space homotopy equivalent to a locally compact ANR (such as a manifold, or a locally finite CW complex) which we denote $\bar X$. Suppose $X$ is homotopy equivalent to a compact space. Then there is a homotopic to the identity map $\bar X\to \bar X$ whose image lies in a compact set. By a standard argument (see e.g. proposition 3.18 of http://arxiv.org/abs/1210.6741 of Guilbault's survey), this is equivalent to assuming that $\bar X$ is finitely dominated (by a finite CW complex). Note that finitely dominated spaces have finitely generated homology groups, and finitely presented fundamental groups (see e.g proposition 3.16 in the above paper). Of course, the fundamental group of a wedge of infinitely many circles is not finitely presented.
Conversely, Ferry proved in "Homotopy, simple homotopy, and compacta" (Topology 9 (1980) pp 101-110) that any space that is dominated by a compact Hausdorff space is homotopy equivalent to a compact Hausdorff space. Thus we conclude that a locally compact ANR is finitely dominated if and only if it is homotopy equivalent to a compact Hausdorff space.
In particular, if $X$ is homotopy equivalent to a smooth manifold that is not finitely dominated, then $X$ is not homotopy equivalent to a compact Hausdorff space, even though any vector bundle over $X$ is a summand of the trivial bundle.
answered Nov 29, 2013 at 22:08
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1$\begingroup$ Why does $f(K)$ have to lie in a subcomplex? On the surface this is simply wrong, as you could have a space filling curve or other exotic phenomena. You probably mean up to homotopy, but it's still not obvious to me since $K$ is not assumed to be a cell complex. $\endgroup$ Commented Nov 29, 2013 at 22:16
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1$\begingroup$ @JimConant, compact subsets of a CW-complex are contained in a finite subcomplex. $\endgroup$ Commented Nov 29, 2013 at 22:22
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$\begingroup$ @MarianoSuárez-Alvarez: but how do we know that the image lies in a CW complex? $K$ is not assumed to be a CW complex here. $\endgroup$ Commented Nov 29, 2013 at 22:31
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2$\begingroup$ It depends on the CW-structure you start with, of course. If you build $X$ using one $1$-cell per circle, then there is no choice. $\endgroup$ Commented Nov 29, 2013 at 22:38
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1$\begingroup$ The problem is that I had X and K reversed in my head. I now fully agree with the argument! $\endgroup$ Commented Nov 29, 2013 at 22:45