Composition of fibrations, fiber homotopy equivalence?

Question

Let $p: D \to B$ and $q: E \to B$ be fibrations and let $f: D \to E$ be a map such that $q \circ f = p$. If $f$ is a homotopy equivalence, does it necessarily follow that $f$ is a fiber homotopy equivalence?

Answer 1

There is a more general result here. Suppose given a commutative diagram of maps

$$\begin{matrix} D & \xrightarrow{f} & E \\ p \downarrow && \downarrow q \\ A & \xrightarrow{g}& B \end{matrix} $$ such that $p,q$ are fibrations and $f,g$ are homotopy equivalencies. let $k$ be any homotopy inverse of $g$, and let $H:gk \simeq 1: B \to B,\, K:kg \simeq 1: A \to A $ be homotopies. Then there is a homotopy inverse $l$ of $f$ such that $pl=kq$ and there is a homotopy $K': lf \simeq 1$ which covers $K$ and a homotopy $H': fl \simeq 1$ which covers a kind of "conjugate homotopy" $$K + k H(g \times 1) - K(kg \times 1) . $$ In particular, the pair $(f,g)$ is a fibre homotopy equivalence. Not that if $A=B$ and $g=1$ then we can take the homotopies $H,K$ to be constant.

This result with different notation is Theorem 3.4 in the paper

R. Brown and P.R.Heath, ``Coglueing homotopy equivalences'', Math. Z. 113 (1970) 313-362.

available here. The main result of the paper is the dual of a glueing theorem for homotopy equivalences proved in all editions of Topology and Groupoids, Chapter 7.