I'm having some trouble with one part of Bott–Tu Theorem 6.8, which proves that homotopic maps induce isomorphic vector bundles under pullback.

The setup is that $\pi:Y\times I\to Y$ is the projection, $f:Y\times I\to X$ is a homotopy between $f_0$ and $f_1$, and $E$ is a vector bundle over $X$. Let $f_{t_0}^{-1}E$ be isomorphic to the vector bundle $F$.

Now there are bundles $f^{-1}E$ and $\pi^{-1}F$ over $Y\times I$. We also have a section of $\text{Hom}(f^{-1}E,\pi^{-1}F)$ over $Y\times t_0$, where the bundles are isomorphic. Covering $Y\times\{t_0\}$ with finitely many trivializing open sets for the Hom-bundle (we assume $Y$ is compact), the authors then say the following: "As the fibers of $\text{Hom}(f^{-1}E,\pi^{-1}F)$ are Euclidean spaces, the section over $Y\times t_0$ may be extended to a section of $\text{Hom}(f^{-1}E,\pi^{-1}F)$ over the union of these open sets."

I don't understand why this is true, and why this is dependent on the fibers being Euclidean (as opposed to $\text{Hom}(f^{-1}E,\pi^{-1}F)$ being a general fiber bundle). I feel that I am missing something really obvious here.

  • $\begingroup$ I think it may just be the sort of argument discussed here: math.stackexchange.com/questions/1213987/…, but I cannot immediately intuit why this would fail for a general fiber bundle. $\endgroup$ Dec 12, 2022 at 20:29
  • $\begingroup$ @TabesBridges Hmm okay so I think that argument does work (and it fails for fiber bundles since it makes use of local frames), though it's now not very clear to me why it was important that we cover $Y\times t_0$ with only finitely many such open sets. It seems as though this argument should hold either way... (using a partitions of unity argument) $\endgroup$
    – boink
    Dec 12, 2022 at 21:01
  • $\begingroup$ The linked argument shows that section can be extended. The compactness/finiteness is relevant cause it implies that the neighborhood to which the section can be extended contains an open set of the form $Y\times J$ with $J$ an open interval about $t_0$ (this is essentially the tube lemma). And an isomorphism $f_t^{\ast}E\cong F$ is equivalent to a section along $Y\times\{t\}$, so this is the condition you want. $\endgroup$
    – Thorgott
    Dec 13, 2022 at 16:01
  • $\begingroup$ @Thorgott Oh right, I forgot about that part of the argument. Thanks! $\endgroup$
    – boink
    Dec 15, 2022 at 0:16


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