Let $p:E\to B$ and $p':E'\to B$ be fibrations. It is [well known](https://en.wikipedia.org/wiki/Fiber-homotopy_equivalence) 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?
> 
> 
> 



fibre-preserving homotopy equivalence
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.)


