I am trying to show that if $p:E\to B$ is a fibration and $f:X\to B$ a homotopy equivalence, then the map $\pi_2:X\times_B E\to E$ is a homotopy equivalence.

Here are my thoughts: Since $f$ is a homotopy equivalence, $f$ has a homotopy inverse $g:B\to X$ such that $g\circ f\simeq id_{X}$ and $f\circ g \simeq id_{B}$. Then I have tried to use the fact that since $X\times_B E$ is a pullback, there exists map $q:X\to X\times_B E$. My guess for the homotopy inverse of $\pi_2$ is the composition $k=q\circ g \circ p$. I am stuck showing that $\pi_2\circ k \simeq id_E$ and $k\circ \pi_2\simeq id_{X\times_B E}$. I think this is because I haven't used the fact that $p$ is a fibration.

  • $\begingroup$ See math.stackexchange.com/questions/83644/… $\endgroup$ – Ronnie Brown Mar 4 '17 at 8:05
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    $\begingroup$ The dual result is in "Topology and Groupoids" Section 7.5. You need to bring in operations of groupoids on certain homotopy classes of maps - analogously to the proof that for well pointed spaces a homotopy equivalence is also a pointed homotopy equivalence. $\endgroup$ – Ronnie Brown Mar 4 '17 at 17:25

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