I am working on this problem and I cant seems to go any further with this problem. Any help will be greatly appreciated.


Let $p:E \rightarrow B$ be a fibration. Suppose that we have two lifts say $H,\hat{H}: \mathbb{I} \times X \rightarrow E$. How can we show that these two homotopies are vertically homotopic? My understanding of this is that lets say we have the following diagram with lifts $H$ and $\hat{H}$ $$\begin{equation*}\begin{array}{ccl} \{0\} \times X&\stackrel{f}{\longrightarrow}&E\\ \!\!\!\!\!\!\!\!{\scriptstyle}\downarrow&&\downarrow{\scriptstyle p}\\ \!\!\!\!\mathbb{I} \times X&\stackrel{h}{\longrightarrow}&B \end{array}\qquad\qquad\end{equation*}.$$

Then we need to show that there exists a homotopy $\phi: \mathbb{I} \times \mathbb{I} \times X \rightarrow E$ with $\phi(0,-,-)= H$ and $\phi(1,-,-)= \hat{H}$ and $p\phi(s,t,x) =h(t,x) \quad \forall (s,t,x) \in \mathbb{I} \times \mathbb{I} \times X $. Any help on how to show this?


First consider the diagram $\require{AMScd}$ \begin{CD} \{0,1\} \times I \times X \cup (I \times (\{0 \} \times X)) @>\varphi>> B \\ @VVV @VVpV\\ I \times (I \times X) @>>h> B \end{CD} where $\varphi(0,t,x)= H(t,x)$, $\varphi(1,t,x)= \hat{H}(t,x)$, $\varphi(s,0,x)= f(x)$

Now let $V= \{0,1\} \times I \cup I \times \{0\} \subset I \times I$ then there exists a homeomorphism(draw the square and $V$) $\Phi: I^2 \to I^2$ such that $\Phi(V)= \{0\} \times I$. Then we have the following diagram \begin{CD} \{0 \} \times I \times X @<\Phi \times id<< V \times X @>\varphi>> E \\ @VVV @VVV @VVpV \\ I \times I \times X @<\Phi<< I^2 \times X @>h>> B \end{CD} Since $p :E \to B$ is a fibration, we can apply the homotopy lifting property to the outer square and get a lift $\phi : I \times I \times X \to E$ so that $\phi(0,-,-) =H$, $\phi(1,-,-)= \hat{H}$ and $p\phi(s,t,x)= h(t,x)$.

  • $\begingroup$ You are welcome. $\endgroup$ – 512122 Mar 20 '17 at 3:04

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