I am reading "Modern Classical Homotopy Theory" by Strom and have come across the following. We are given a fibration $F\rightarrow E\rightarrow B$. One then has two pushout squares: $$\require{AMScd} \begin{CD} F @>>> CF\\ @VVV @VVV \\ E @>>> E\cup CF \end{CD} $$ and $$\require{AMScd} \begin{CD} * @>>> *\\ @VVV @VVV\\ B @= B \end{CD} $$ There are clear maps between these squares, resulting in a cube (that I can't figure out how to draw using MathJax). This gives a map between the pushouts $E\cup CF\rightarrow B$. This came about when examining another question. There I have $$ F\stackrel{i}{\longrightarrow} E\stackrel{p}{\longrightarrow} B\stackrel{f}{\longrightarrow K}, $$ where $E$ is the homotopy fiber of $f$ and $F$ is the homotopy fiber of $p$.

I am just hoping for someone to help clear up some of my confusion. The composite $$ E\cup CF\longrightarrow B\longrightarrow K $$ needn't be nullhomotopic. But it would seem from the pushout cube and the fact that $E$ is the fiber of $f$, that it must be. Can someone explain why not?


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