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.
Best Answer
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.