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?



How does all of the bundles over a certain manifold characterize the homotopy class of the base manifold?
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.


--------Edit--------


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.


