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I'm trying to understand the proof of Theorem 2.1 in "The Topology of Fiber Bundles" found online at http://math.stanford.edu/~ralph/fiber.pdf. enter image description here

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What I don't understand is how do we actually define $\tilde{H}$ and the claim that "This is clearly a bundle isomorphism since it induces the identity map on both the base space and on the fibers".

I'm not sure what does "induces the identity on fibers" means? hence I don't know how to check that $\tilde{H}$ map does induces the identity on fibers. And I don't know why such condition actually implies isomorphism.

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  • $\begingroup$ Please accept the answer if you find it satisfactory. $\endgroup$
    – Arrow
    May 10, 2019 at 10:32

1 Answer 1

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Consider the bundles $f_0^*(E)\to X$ and $H^*(E)\to X \times I$. By the theorem which is mentioned you get the bundle map $$\begin{array}{c}f_0^*(E) \times I &\to &E\\ \downarrow && \downarrow \\ B\times I &\to & B\end{array} $$

By the universal property of a pullback, the given maps give us a bundle map to $H^*(E)$: $$\begin{array}{c}f_0^*(E) \times I &\to &H^*E\\ \downarrow && \downarrow \\ B\times I &\stackrel {id} \to & B \times I\end{array} $$

But by restricting you get that this is a bundle isomorphism. But by restricting to $B\times 1$ you also get a bundle isomorphism $f_0^*E \to f_1^*E$.

Let me know if you would like to have further assistance.

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  • $\begingroup$ Thank you for your reply. The only point in the proof that I don't understand is why is the map $f^*_0(E) \times I \rightarrow H^*E$ an isomorphism. If you could expand the reasoning that would be great. $\endgroup$
    – user113988
    Dec 15, 2014 at 13:38
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    $\begingroup$ Because $f_0^*E\times I\to E$ is fibrewise an isomorphism, since $f_0^*E \to E$ is one, as well as $H^*E \to E$ is fibrewise an isomorphism. Hence $f_0^*E \times I \to H^*E \to E$ is fibrewise an iso, the latter map in the composition is fibrewise an iso, hence also the former. So you are left to show that $id$ is an isomorphism of topological spaces. Is it alright now? $\endgroup$ Dec 16, 2014 at 9:05

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