fibre-preserving homotopy equivalence

Question

Let $p:E\to B$ and $p':E'\to B$ be fibrations. It is well known that if $f:E\to E'$ a fibrewise map that is also a homotopy equivalence, then it is a fibrewise homotopy equivalence.

What about the more general situation of fibrations $p:E\to B$ and $p':E'\to B'$ over different bases? A fibre-preserving map from $p$ to $p'$ is a pair of maps $f:E\to E'$ and $\overline{f}:B\to B'$ such that $\overline{f}\circ p=p'\circ f$, and a fibre-preserving homotopy between such maps is a pair of homotopies $H:E\times I\to E'$ and $\overline{H}:B\times I\to B'$ such that the following diagram commutes: $\require{AMScd}$ \begin{CD} E\times I @> H >> E'\\ @V p\times\operatorname{Id} V V @VVp'V\\ B\times I @> \overline{H}>> B' \end{CD}

One can then easily define a fibre-preserving homotopy equivalence from $p$ to $p'$.

Question: Is there an example of a fibre-preserving map of fibrations \begin{CD} E @> f >> E'\\ @V p V V @VVp'V\\ B @> \overline{f}>> B' \end{CD} such that $f$ and $\overline{f}$ are homotopy equivalences, but the pair $(f,\overline{f})$ is not a fibre-preserving homotopy equivalence, i.e. does not admit a fibre-preserving homotopy inverse?

Answer 1

The answer to the question as asked is no: a fibre-preserving map of fibrations in which the maps of total and base spaces are homotopy equivalences is neccessarily a fibre-preserving homotopy equivalence (also known as a homotopy equivalence of fibrations). A reference was supplied in the comments by Gustavo Granja, to Peter May's book A Concise Course in Algebraic Topology, where the statement appears as a Proposition on page 53. (The proof, although not given in detail, does appear to be a straightforward dualization of the corresponding result for cofibrations, proved on page 48.)