Fiber homotopy equivalence

Question

Suppose $(E_1, B_1, p_1)$ and $(E_2, B_2, p_2)$ are homotopy equivalent Hurewicz fibrations (that is there exist homotopy equivalencies $\phi: E_1 \rightarrow E_2$ and $\psi: B_1 \rightarrow B_2$ such that $p_2 \circ \phi \sim \psi \circ p_1$). Is it true that the fibers of those fibrations are homotopy equivalent?

Answer 1

By taking the pullback in your square, you may assume that $B_1 = B_2$. Then your question becomes: are the "equivalent" fibrations fiber homotopy equivalent?

The answer to this is yes: you can find it at the very beginning of Chapter 5 in May's "A concise course..." (p. 52); or indeed, the result you want is the next proposition on the following page.

Answer 2

This is mainly a special case of the main result of

[1] R. Brown and P.R. Heath, "Coglueing Homotopy Equivalences", Math. Z. 113, 313-325 (1970),

which in brief terms says that under suitable fibration conditions a pullback of homotopy equivalences is a homotopy equivalence. This is now seen as result in abstract homotopy theory,but an advantages of the methods of this paper is that there is control of the homotopies involved (see Theorem 3.4).

Actually since you assume only $p_2\circ \phi \sim \psi \circ p_1$ there is needed a further step, that a pullback $f^*(E)$ of a fibration $p:E \to B$ depends, up to homotopy equivalence, only on the homotopy class of $f$. This theme is pursued in further papers of Heath under the title of "Groupoid operations and fibre-homotopy equivalence".