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Let $E \to M$ be a fiber bundle and $f,g \colon N \to M$.

  • We know (e.g. from Steerod) that if $f$ and $g$ are homotopic, then the pullback bundles are equivalent bundles over $N$, $f^\ast E \cong g^\ast E$.
  • What about the reverse implication? If we assume that $f^\ast E \cong g^\ast E$, in what cases does this imply that $f$ and $g$ are homotopic?
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    $\begingroup$ Surely letting $E$ be the trivial bundle on $N$ gives counter-example? $\endgroup$
    – krm2233
    Nov 30, 2023 at 0:39
  • $\begingroup$ The case where this is naturally true is when $M$ is the classifying space for the respective type of fiber bundle. $\endgroup$
    – Thorgott
    Nov 30, 2023 at 3:06
  • $\begingroup$ Thanks @krm2233, yes I suppose I am looking for general conditions (e.g. $E$ not trivial) such that the implication holds. $\endgroup$
    – Joe
    Nov 30, 2023 at 21:15
  • $\begingroup$ @Thorgott can you expand a bit on that? $\endgroup$
    – Joe
    Nov 30, 2023 at 21:15
  • $\begingroup$ The classifying $BG$ is equipped with a universal bundle such that every fiber bundle with structure group $G$ on a space $M$ is (under mild technical conditions) the pullback of this universal bundle by a unique homotopy class of maps $M\rightarrow BG$. More generally, using this, the property you want to hold for all $N$ is equivalent to asking that the map $M\rightarrow BG$ classifying $E$ is a monomorphism in the homotopy category. These are incredibly hard to understand. $\endgroup$
    – Thorgott
    Nov 30, 2023 at 22:18

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