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I think this question is related to what is known as "obstruction theory", but I'm not very familiar with this field of mathematics, so I am asking here.

Let $\pi:N\rightarrow M$ be a smooth fibered manifold and $\xi:E\rightarrow N$ a smooth vector bundle over its total space. Suppose furthermore that $\pi:N\rightarrow M$ is a homotopy equivalence. If this is not sufficient, it may also be assumed that $\pi$ is an affine bundle or a composition of affine bundles.

Claim 1: There is a vector bundle atlas $(U_\alpha,\varphi_\alpha)$ of $\xi:E\rightarrow N$ such that each set $U_\alpha\subseteq N$ is "cylindrical" with respect to the fibration $\pi$, meaning that there is an open set $U_{\alpha,0}\subseteq M$ with $U_\alpha=(\pi^{-1})(U_{\alpha,0})$.

Essentially, I'd like to know whether Claim 1 is true or not.


The question is self-contained as given above, but I am adding some background. I am particularly interested in sufficiency conditions for a vector bundle on $N$ to be the pullback of a vector bundle on $M$.

I didn't go though the reasoning very rigorously, but I think that if Claim 1 is true, I can prove via a Swan--Serre type reasoning that

Claim 2: If $\xi:E\rightarrow N$ is a vector bundle and $\pi:N\rightarrow M$ is a homotopy equivalence, then there is a vector bundle $\xi^\prime:E^\prime\rightarrow M$ such that $(E,\xi,N)\cong\pi^\ast(E^\prime,\xi^\prime, M)$.

If it happens that I am dead wrong with this, then I am also interested in conditions where this holds.


Finally, the actual motivation is to establish a Swan--Serre-type theorem for infinite jet bundles.

Claim 3: Let $J^\infty(\pi)$ be the infinite jet bundle of $\pi:N\rightarrow M$, let $A:=\mathcal F_\infty(\pi)$ be the commutative ring of smooth functions on $J^\infty(\pi)$ (interpreted as the direct limit of rings of smooth functions on each finite order jet bundle). If $P$ is a projective, finitely generated $A$-module then there is a vector bundle $\xi:E\rightarrow N$ such that $P\cong\Gamma_\infty(\xi)$, where $\Gamma_\infty(\xi)$ is the $A$-module of smooth generalized sections of $\xi$, that is the set of all smooth maps $\gamma:J^\infty(\pi)\rightarrow E$ satisfying $\xi\circ\gamma=\pi^\infty_0$.

Similar to Claim 2 I only have a proof-draft for this, but I think if Claim 2 is not absolutely necessary for this, then it is extremely helpful.


TL;DR: I am mainly interested in whether Claim 1 is true or false, but any additional info on the validities of Claim 2 or Claim 3 is also welcome.

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  • $\begingroup$ Isn't it valid that we take the homotopy inverse of $\pi$ and pull the bundle $E$ back along it to get $E'$ in Claim $2$? $\endgroup$
    – Zerox
    Commented Mar 23 at 17:38
  • $\begingroup$ @Zerox Yeah I'll need to check this more explicitly but I agree and I think Claim 2 is way more trivial than I thought lol. $\endgroup$ Commented Mar 23 at 17:39

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As indicated in the comments, this question ended up being accidentally rather trivial.

Specifically, the following three facts are fairly well-known and rather easy to establish:

  1. For homotopic smooth maps $f,g:M\rightarrow N$ and a vector bundle $\xi:E\rightarrow N$ we have $f^\ast(\xi)\cong g^\ast(\xi)$.
  2. For smooth maps $g:M\rightarrow N$, $f:N\rightarrow P$ and a vector bundle $\xi:E\rightarrow P$ we have $(f\circ g)^\ast(\xi)\cong g^\ast f^\ast(\xi)$.
  3. $(\mathrm{id}_M)^\ast(\xi)\cong\xi$.

Thus if $\phi:M\rightarrow N$ is a homotopy inverse for $\pi:N\rightarrow M$, we set $\xi^\prime:=\phi^\ast(\xi)$. Then $$ \pi^\ast(\xi^\prime)\cong(\phi\circ\pi)^\ast(\xi)\cong(\mathrm{id}_N)^\ast(\xi)\cong\xi. $$

This proves Claim 2. Claim 1 then follows from Claim 2. as we can trivialize $\xi$ by trivializing $\xi^\prime$ over a cover of $M$.

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  • $\begingroup$ Of course in 1. you want to assume f and g are homotopic. $\endgroup$
    – Dan Ramras
    Commented Mar 23 at 20:14
  • $\begingroup$ @DanRamras Thanks for the comment, fixed it. $\endgroup$ Commented Mar 23 at 20:32