It is known that if $f:M\rightarrow N$ is a homotopy equivalent, then the the process of pullback gives a one-one correspondence between bundles over $N$ and $M$ up to isomorphism. Is the converse( that if $f$ gives a 1-1 correspondence between them, then $f$ is a homotopy equivalence)true? Or any counterexample?

By the way, are bundles of a fixed rank over a compact manifold finite up to isomorphism? And is there any characterization of them like the holomorphic line bundles as a first cohomology group of a certain sheaf? If so, is there any way to calculate?

$M,N$ are smooth manifolds, and I originally thought of real vector bundles. But since this is a wild guess, it might be better to think the most general bundles you could imagine that homotopy invariance holds, even though I only know the real case.

It seems algori's answer apply to all kind of vector bundles I can imagine, though a bit beyond me. Thanx very much. And any reference for your arguments?

And I'd like to know whether homotopy invariance holds in fiber bundles.

Thanks all of you. But formally whose answer should I accept?

  • $\begingroup$ What do you mean in the last question by "can we determine?"? In what sense? We can considering the set of all bundles, for example... is that "determining" them? $\endgroup$ – Mariano Suárez-Álvarez Apr 20 '11 at 4:34
  • $\begingroup$ There are many counter-examples. All orientable 3-manifolds have trivial tangent bundles, so all maps between them induce isomorphisms of their tangent bundles, yet most such maps are not homotopy-equivalences. $\endgroup$ – Ryan Budney Apr 20 '11 at 4:36
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    $\begingroup$ Ryan, that sounds different from what lethe was asking. His condition is stronger, for example, than requiring that $f$ induce an isomorphism on K-theory. $\endgroup$ – Dan Ramras Apr 20 '11 at 4:45
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    $\begingroup$ @lethe : The set of isomorphism classes of bundles of a rank $k$ on $X$ is equal to $H^1(X;F)$, where $F$ is the sheaf of continuous functions with values in $GL_k(\mathbb{R})$. $\endgroup$ – Andy Putman Apr 20 '11 at 4:47
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    $\begingroup$ @ lethe - That is part of Steenrod's thesis. There's a quick proof (for closed manifolds) via differential cohomology machinery - $M$ is orientable means $w_1(TM)=0$. Since $M$ is odd dimensional, $\chi(M)=0$ which implies $w_3(TM)=0$. It is known that the total Stiefel-Whitney class $w=1+w_1+w_2+w_3$ is the Steenrod square of $v=1+v_1+v_2+v_3$, called the Wu class. Using this one can show that $w_2(TM)=0$. Since $w_2$ is the obstruction to the existence of a $2$-frame (which gives $2$ lin. ind. nowhere zero vector fields) we can use the orientation frame to get $3$ such vector fields. $\endgroup$ – Somnath Basu Apr 20 '11 at 5:29

If you are liberal enough as to what you consider to be "a bundle", then this is true. The reason is that the symmetric product $$ G := SP^\infty(S^{n-1}) \simeq K(\mathbb{Z},n-1)$$ is a topological abelian group, and principal $G$-bundles are classified by maps to $BG \simeq K(\mathbb{Z}, n)$. Thus if $f : X \to Y$ induces a bijection between all fibre bundles, then it is a (co)homology equivalence.

Considering all principal $H$-bundles for $H$ a discrete group, one also sees that $f$ induces an isomorphism on $\pi_1$ (as $f_* : \mathrm{Hom}(\pi_1(X), H) \cong \mathrm{Hom}(\pi_1(Y), H)$ for all groups $H$ means $\pi_1(X) \to \pi_1(Y)$ is an isomorphism).

Thus $f$ has to be a weak homotopy equivalence. As the question as asked for compact smooth manifolds, which are homotopy equivalent to CW-complexes, $f$ is a homotopy equivalence.


As TG points out in the comments, $\pi_1$ equivalence and homology equivalence are not enough to deduce weak homotopy equivalence without a simpleness assumption.

Thus: for each discrete group $H$ and $H$-module $V$ we can form $$G := K(V, n-1) \rtimes H$$ a topological group. A concrete model for $K(V, n-1)$ can be taken to be $SP^\infty_V(S^{n-1})$, the symmetric product of the sphere with labels in $V$, on which $H$ acts on the labels. A principal $G$-bundle over $X$ is classified by a map to the total space of the fibration $$K(V, n) \to BG \to BH,$$ and this consists of a homomorphism $f:\pi_1(X) \to H$ and a twisted cohomology class in $H^n(X; f^*V)$.

If I am not mistaken again, a map inducing a $\pi_1$-iso and an iso on cohomology with all local coefficients is indeed a weak equivalence.

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    $\begingroup$ Or instead of using topological groups that are either abelian or discrete, you can use more general ones and say that $M$ and $N$ themselves are equivalent to classifying spaces, and do it that way. This is if they are connected. If they are not, then it is false. $\endgroup$ – Tom Goodwillie Apr 20 '11 at 13:55
  • $\begingroup$ They won't typically be equivalent to classifying spaces of actual groups, though. $\endgroup$ – Oscar Randal-Williams Apr 20 '11 at 14:09
  • $\begingroup$ You mean, because the loopspace is not quite a group? but there are ways of rectifying that, for example Kan's Quillen adjunction between simplicial groups and reduced simplicial sets. $\endgroup$ – Tom Goodwillie Apr 20 '11 at 14:35
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    $\begingroup$ Wait a minute: The relative Hurewicz Theorem is not that strong. A map inducing isomorphisms on $\pi_1$ and $H_n$ for all $n$ need not be a weak equivalence. $\endgroup$ – Tom Goodwillie Apr 21 '11 at 2:38
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    $\begingroup$ You'd need the space to be simple (i.e. the action of $\pi_1$ to be trivial on higher homotopy groups) in order to apply the homology version of Whitehead here. $\endgroup$ – Dylan Wilson Apr 21 '11 at 3:37

It all depends on which bundles we are considering. If we take topological vector bundles with structure group a Lie group $G$, then there aren't any non-trivial bundles over the 3-sphere: we have $\pi_3(BG)=\pi_2(G_0)=0$ where $G_0$ is the connected component of the unit of $G$.

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    $\begingroup$ Great answer! Why do you specify topological, and not smooth, vector bundles? $\endgroup$ – Mark Grant Apr 20 '11 at 6:01
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    $\begingroup$ This true if $G$ is a finite dimensional Lie group. Loop groups give simple counterexamples otherwise. But then most vector bundles are of finite rank ;-) $\endgroup$ – David Roberts Apr 20 '11 at 6:14
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    $\begingroup$ David -- those disclaimers never help, do they;)? $\endgroup$ – algori Apr 20 '11 at 6:21
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    $\begingroup$ By the way, there do exist nontrivial smooth sphere bundles over $S^3$. Their structure group of course is not reducible to a Lie group. This boils down to finding a sphere whose diffeomorphism group has nontrivial $\pi_2$. I think, examples can be found be in [Antonelli- Burghelea-Kahn, "The non-finite homotopy type of some diffeomorphism groups", Topology 11 (1972), 1–49]. $\endgroup$ – Igor Belegradek Apr 20 '11 at 12:07
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    $\begingroup$ If lethe would prefer an example $M\to N$ where $M$ and $N$ are closed manifolds of the same dimension (rather than $S^3$ and a point) then there is also $S^3\times S^3\to S\wedge S^3=S^6$. $\endgroup$ – Tom Goodwillie Apr 20 '11 at 12:25

Let me attempt to summarise some basic facts.

Let $G$ be a Lie (say) group, and let's let $M$ and $N$ be smooth compact manifolds as in the question. A principal $G$-bundle on $N$ is a bundle with structure group $G$ and fibre $G$, and transition functions given by left translation. Alternatively, a principal $G$-bundle on $N$ is a free $G$-space $\tilde{N}$ with orbit space $N$. Principal $G$ bundles on $N$ have a nice homotopy classification. Namely, the set of isomorphism classes of principal $G$-bundles is in one-to-one correspondence with the set of homotopy classes $[N,BG]$, where $BG$ is a classifying space of the group $G$.

Not only that, the classification is functorial. Given a principal $G$-bundle $\xi$ on $N$ and a map $f\colon M\to N$, the pullback construction gives you a bundle $f^{\ast}\xi$ on $M$, and the corresponding operation on homotopy classes $f^{\ast}\colon [N,BG]\to [M,BG]$ is simply pre-composition with the homotopy class of $f$. This implies that a homotopy equivalence $f\colon M\to N$ gives a one-to-one correspondence between isomorphism classes of principal $G$-bundles over $N$ and $M$.

Now let $G$ be a subgroup of $Gl(n,\mathbb{R})$. It turns out that rank $n$ vector bundles with structure group $G$ are pretty closely related to principal $G$-bundles. In fact the isomorphism classes of these two objects correspond. You can read about this in these notes (see Proposition 4.1) or in the classic book reference "Fibre bundles" by Dale Husemoller.

Two special cases relevant to your second paragraph:

  • If $G=O(1)$ then we are classifying real line bundles. Since $O(1)\cong \mathbb{Z}_2$ we have $BO(1)=K(\mathbb{Z}_2,1)=\mathbb{R}P^{\infty}$, and real line bundles on $M$ are classified by elements of $[M,K(\mathbb{Z}_2,1)]\cong H^1(M;\mathbb{Z}_2)$.

Since you assume $M$ is compact, this will imply that there are finitely many real line bundles on $M$ (up to isomorphism).

  • If $G=U(1)$ then $BU(1)=K(\mathbb{Z},2)=\mathbb{C}P^{\infty}$ and complex line bundles are classified by $[M,K(\mathbb{Z},2)]\cong H^2(M;\mathbb{Z})$.

Thus for any compact manifold $M$ with nonzero second Betti number, there are infinitely many non-isomorphic complex line bundles over $M$.

A similar homotopy classification holds for bundles with more general fibres. In particular, given a $G$-space $F$ and a principal $G$-bundle, one can build a fibre bundle with fibre $F$ and structure group $G$. Also you can go the other way; each fibre bundle has an underlying principal bundle (see Husemoller chapter 4). So the answer to your final question is yes.

  • $\begingroup$ Thanks very much! I start to figure out why all of these answers work. Since all of the answers are of great help, I'm quite uncertain whose answer to accept, though it's merely formal. $\endgroup$ – Honglu Apr 21 '11 at 1:52
  • $\begingroup$ 1. I think I misunderstood the real question - if this was whether a map which induces equivalences of all fibre bundles is a homotopy equivalence, then Oscar seems to have come up with a good answer! 2. I wish MO had been around when I was your age :) $\endgroup$ – Mark Grant Apr 21 '11 at 14:52

All the above answers are great, but as you are just learning this stuff, the following might be helpful. Vector/$G$-bundles contain actual geometric information! The tangent bundle, for example, tells you a lot about the geometric structure of the underlying manifold (some definitions of curvature etc.). Now a homotopy theorist can measure these things using cohomology and homotopy as described above by Algori and Mark Grant. So cohomology restricts what kind of bundles you can have (are they trivial? how many linearly independent sections might they have?). Chern-Weil theory might be something to look up if this sounds cool.

Something to note though, all vector bundles are homotopy equivalences.

Also, Peter May wrote a little memoir called Classifying spaces and fibrations where he writes down classifying spaces for fibrations with fiber $F$. These have significantly less structure than a plain old fiber bundle. Granted, there are restrictions on when we have classification, but usually they are pretty harmless, and any compact smooth manifold should satisfy them.


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