$\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ I am trying to show that for a given fibration $E \xrightarrow{p} B$, and a cell structure $\{B_n\}$ on $B$ that , that $p^{-1}B_n \to p^{-1}B_{n+1}$ is a cofibration. I am trying to deduce it from a more general fact(that I hope is true).
Question:
Let $E_1$ be the pullback of a fibration $E_2 \xrightarrow{p_2} B_2$ along a cofibration $j$:
$ \begin{CD} E_1 @>i>> E_2 \\ @V\text{fibration}Vp_1V @Vp_2V\text{fibration}V\\ B_1 @>cofibration>j> B_2 \\ \end{CD}$,
Does it follow that $i$ is a cofibration?
Note: This question comes from my previous quetsion Partial Converse to "Pushout of a cofibration is a cofibration". I did not realize that I needed to further specify that the above square be a pullback square so I created this question.
Without loss of generality assume that $B_1 \to B_2$ is an inclusion(This is satisfied for the cofibration $B_p \to B_{p+1}$ that you are interested in). The only thing I need to use is $B_1 \to B_2$ is a cofibration $\iff$ $(B_2,B_1)$ an NDR Pair. Suppose that its given by the homotopy $h: B_2 \times I \to B_2$ and $u: B_2 \to I$.
For the obvious lift $E_2 \times 0 \to E_2$ $\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ the diagram $ \begin{CD} E_2 \times 0 @>>> E_2 \\ @VVV @Vp_2VV\\ E_2 \times I @>(h:B_2 \times I \to B_2)\circ p_2>> B_2 \\ \end{CD}$, has a solution $H: E_2 \times I \to E_2$. Its a 5 second mental exercise to check that $(E_2,E_1)$ is an NDR pair given explicitly by the homotopy $H$ and $u \circ p: E_2 \to I$. Therefore $E_1 \hookrightarrow E_2$ is a cofibration.