I'm actually studying obstruction theory as presented in the last section of chapter $4$ of the book Algebraic Topology by Allen Hatcher. He first finds condition so that a space $X$ admits a Postnikov tower of principal fibrations, which is a Postnikov tower

$\require{AMScd}$ \begin{CD} \vdots & & \vdots\\ @| & @VVV \\ X @>>> X_3 \\ @| & @VVV \\ X @>>> X_2\\ @| & @VVV\\ X @>>> X_1 \end{CD} where any map $ X_n \rightarrow X_{n-1}$ is such that exists a fibration sequence $F\rightarrow E\rightarrow B$ and weak homotopy equivalences $X_n\rightarrow F$ and $X_{n-1}\rightarrow E$ such that

$\require{AMScd}$ \begin{CD} X_n @>>> X_{n-1} \\ @VVV & @VVV \\ F @>>> E@>>> B\\ \end{CD} commutes.
The problem is that when a few pages later he deals with the problem of finding a map $W\rightarrow X_n$ to complete the commutative diagram $\require{AMScd}$ \begin{CD} A @>>> X_n \\ @VVV & @VVV \\ W @>>> X_{n-1}\\ \end{CD} where $(W,A)$ is a $CW$ pair and $A\rightarrow W$ is the inclusion, he supposes that $X_n$ actually is the pullback of the path fibration $PB\rightarrow B$ via $X_{n-1}\rightarrow B$. Why can he suppose this?

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    $\begingroup$ The assertion is basically that if $E\to B$ is a fibration with fiber $F$, and if $PB\to B$ is a fibration such that the space $PB$ is contractible, then the space $E\times_BPB$ is equivalent to $F$. This is so because it fibers over the contractible space $PB$ with fiber $F$. $\endgroup$ Oct 6, 2018 at 22:45
  • $\begingroup$ That's ok but it seems to me that in this way, trying to adapt what the Hatcher does, I can only deduce a map that makes the diagram commute up to homotopy. Anyway I need it to strictly commute, and even using the fact that the map $A\rightarrow W$ is a cofibration and the map $X_n\rightarrow X_{n-1}$ is a fibration it's not clear to me how to obtain a map $W\rightarrow X_n$ that makes the diagram commute. $\endgroup$
    – Diego95
    Oct 7, 2018 at 0:36
  • $\begingroup$ If $A\to W$ is a cofibration and $X_n\to X_{n-1}$ is a fibration, then a solution up to homotopy should imply a strict solution. $\endgroup$ Oct 7, 2018 at 18:13
  • $\begingroup$ That's exactly what I don't know how to show. $\endgroup$
    – Diego95
    Oct 7, 2018 at 19:44
  • $\begingroup$ This was also posted in math.stackexchange. $\endgroup$ Sep 22, 2019 at 18:46


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