# Understanding the Homotopy Invariance of Fiber Bundle

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.

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.

• Please accept the answer if you find it satisfactory. May 10, 2019 at 10:32

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.

• 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. Dec 15, 2014 at 13:38
• 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? Dec 16, 2014 at 9:05