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Here is a question about proving the pullback bundles by homotopic maps are isomorphic in Prof. Ralph Cohen notes Bundles, Homotopy, and Manifolds. The proof is in page 73 of the notes. For me, considering only vector bundles is good enough.

The statement is: let $E\to B$ be a vector bundle and $f_0, f_1:X\to B$ are two homotopic maps. Then $f_0^*E\cong f_1^*E$. The proof is attached.

In the proof, the homotopy $\tilde{H}:f_0^*E\times I\to E$, which covers $H:X\times I\to B$, is at the same time a bundle isomorphism. And we get a bundle isomorphism $f_0^*E\to f_1^*E$ by restricting $\tilde{H}$ to $X\times\{1\}$.

My question is: if we restrict $\tilde{H}$ to $X\times\{0\}$, we get a bundle isomorphism $f_0^*E\to f_0^*E$, probably the identity map. Does it mean the bundle isomorphism $\tilde{H}:f_0^*E\times I\to E$ is a homotopy between the bundle isomorphism $f_0^*E\to f_1^*E$ and the bundle isomorphism $f_0^*E\to f_0^*E$? It looks like the answer is yes, but I am not sure.

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  • $\begingroup$ $f_0^*E$ and $f_1^*E$ are two different bundles, for which the theorem gives an isomorphism $f_0^*E\to f_1^*E$. So it's not clear what you mean by a homotopy from the identity on $f_0^*E$ to that isomorphism. What is true is that the isomorphism is determined only up to homotopy (between isomorphisms $f_0^*E \to f_1^*E$). $\endgroup$
    – Steve Costenoble
    Commented Sep 29, 2023 at 16:19
  • $\begingroup$ @SteveCostenoble Sorry, that's a typo. But when you say the isomorphism $f_0^*E\to f_1^*E$ is determined only up to homotopy, do you mean any two isomorphisms $f_0^*E\to f_1^*E$ are homotopic, and moreover, the homotopy is an isomorphism $f_0^*E\times I\to f_1^*E$? $\endgroup$
    – Ho Man-Ho
    Commented Sep 29, 2023 at 17:23
  • $\begingroup$ No, not that any two isomorphisms whatsoever are homotopic. However, $\tilde H$ is only determined up to homotopy (over $H$) and you could vary the homotopy $H$ up to homotopy rel endpoints. The resulting isomorphisms $f_0^*E\to f_1^*E$ will all be homotopic. On the other hand, if $H_1$ and $H_2$ are two homotopies not homotopic rel endpoints, you could end up with two non-homotopic isomorphisms. $\endgroup$
    – Steve Costenoble
    Commented Sep 29, 2023 at 20:30
  • $\begingroup$ And, again, it doesn't make sense to ask that a map $f_0^*E\to f_0^*E$ be homotopic to a map $f_0^*E\to f_1^*E$, since the targets are different bundles. $\endgroup$
    – Steve Costenoble
    Commented Sep 29, 2023 at 20:32
  • $\begingroup$ Another note, sorry: Where you refer to "the bundle isomorphism $\tilde H\colon f_0^*E\times I \to E$," what you really have is an isomorphism $\tilde H\colon f_0^*E\times I\to H^*E$ of bundles over $X\times I$, as in the proof you attached. $\endgroup$
    – Steve Costenoble
    Commented Sep 29, 2023 at 20:56

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