# Questions about obstruction theory (Hatcher's book)

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?

• 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 $F$. – Tom Goodwillie Oct 6 '18 at 22:45
• 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. – Diego95 Oct 7 '18 at 0:36
• 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. – Tom Goodwillie Oct 7 '18 at 18:13
• That's exactly what I don't know how to show. – Diego95 Oct 7 '18 at 19:44
• This was also posted in math.stackexchange. – Arturo Magidin Sep 22 '19 at 18:46