It is well-known that
Suppose that $i:A\to X$ and $i': A\to X'$ are cofibrations and $g:X\to X'$ is a homotopy equivalence so that $g\circ i\simeq i'$ (commutes up to homotopy). Then $g$ is homotopic to a map $h:X\to X'$ so that $h\circ i=i'$.
So, I was wondering if there is a dual version for fibrations:
Suppose that $p:E\to B$ and $p': E'\to B$ are fibrations and $g:E\to E'$ is a homotopy equivalence so that $p'\circ g\simeq p$ (commutes up to homotopy). Then $g$ is homotopic to a map $h:E\to E'$ so that $p'\circ h=p$.
Is this true?
According to the Davis and Kirk (p. 179), the proof of cofibration version mirrors the proof of the theorem states that
any maps of fibrations which is a homotopy equivalence is a fiber homotopy equivalence.
So would this theorem be helpful to get the "theorem" about fibrations?
Any reference or hints are appreciated!
I mean, the proof for the cofibration is a one-line application of the homotopy extension property :
$\require{AMScd} \begin{CD} A @>>> A\times I\\ @VVV @VVV\\ X @>>> X\times I \end{CD}$
with $A\times I \to X'$ is the homotopy from $g\circ i$ to $i'$, and $X\to X'$ is $g$. Then you get a map $H:X\times I \to X'$, and putting $h=H(-,1)$ yields the desired result.
This is all just "formal manipulation", so it must be similar for fibrations, and a one-line application of the homotopy lifting property should work : look at
$\require{AMScd} \begin{CD} E @>>> E'\\ @VVV @VVV\\ E\times I @>>> B \end{CD}$
where the map $E\to E'$ is $g$, the map $E\times I\to B$ is a homotopy from $p'\circ g$ to $p$, so that the diagram commutes if you let $E\to E\times I$ be inclusion at $0$. Then let $H$ be any lift $E\times I\to E'$. $H(-,0) = g$ because the upper triangle commutes, so if you let $h=H(-,1)$, $h$ and $g$ are homotopic.
Looking at the lower triangle, and evaluating at $1$, we get that $p'\circ h = p$