If $p:E_0\rightarrow B_0$ is a fibration, $f:B\rightarrow B_0$ is a homotopy equivalence, then by pullback we get maps $\text{pr}_2:E:=B\times_{B_0}E_0 \rightarrow E_0$ and $\text{pr}_1:E:=B\times_{B_0}E_0 \rightarrow B$. It is well-known that $\text{pr}_1$ must be a fibration. I wish to prove $\text{pr}_2$ is a homotopy equivalence.
I have constructed a map $\ell$ which I suspect to be its homotopy inverse. Unfortunately, when I try to construct a homotopy $\ell \circ \text{pr}_2 \simeq \text{id}_E$, I can only do so "coordinate-wise", i.e. we have $\text{pr}_1 \circ \ell \circ \text{pr}_2 \simeq \text{pr}_1$ and $\text{pr}_2 \circ \ell \circ \text{pr}_2 \simeq \text{pr}_2$. However, these are not enough to prove $\ell \circ \text{pr}_2 \simeq \text{id}_E$ since the resulting homotopy $h:I \times E \rightarrow B \times E_0 \supset E$ might not stay at all times inside $E$ (even though it does stay inside $E$ at times $0$ and $1$).
This is lemma 2.1 in this link: http://www.math.uiuc.edu/~franklan/Math527_0308.pdf But they leave it as an exercise to the reader, although unfortunately I have failed to prove it.
Would anyone have any hints? Thanks!
Thank you for pointing to my notes. The exercise was trickier than I made it sound, so I posted an expanded version. The statement is now Proposition 2.3 here:
http://www.home.uni-osnabrueck.de/mfrankland/Math527/Math527_0308_20170508.pdf