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Homotopy lifting theorem

Let $p:C\rightarrow X$ be a covering map.

Let $F:Y\times[0,1]\rightarrow X$ be a continuous function.

Let $f:Y\rightarrow C$ be a continuous function such that $(p\circ f)(y)=F(y,0)$.

Then, there is a unique continuous $G:Y\times[0,1]\rightarrow C$ such that $G(y,0)=f(y)$ and $(p\circ G)(y,t)=F(y,t)$.

This is the homotopy lifting theorem and I have proved it and understand this theorem.

Meanwhile, below is a theorem in my text.

Let $p:C\rightarrow X$ be a covering map.

Let $F:[0,1]\times [0,1]\rightarrow X$ be a continuous functuon.

Let $e_0\in C$ such that $p(e_0)=F(0,0)$.

Then, there exists a unique continuous $G:[0,1]\times [0,1]\rightarrow C$ such that $(p\circ G)(s,t)=F(s,t)$ and $G(0,0)=e_0$.

How do I apply the homotopy lifting theorem to this one?

Or, should I prove it directly?

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You just need to apply the homothopy lifting thorem twice. Since $p(e_0)=F(0,0)$, by lifting the map $F_0=F \restriction_{\{0\} \times [0,1]}$ you get a map $G_0: \{0\} \times [0,1] \to C$ such that $G_0(0,t)=F(0,t)$ and $G(0,0)=e_0$. Now you can lift the whole map $F$ to a map $G:[0,1] \times [0,1] \to C$ such that $(p \circ G)(s,t)=F(s,t)$ and $G(0,t)=G_0(0,t)$, in particular $G(0,0)=G_0(0,0)=e_0$.

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  • $\begingroup$ @aroval you are constructing $G_0$ as a lift of $F(0,t)$. But to apply lifing theorem, shouldn't we construct a lift of $F(s,0)$? $\endgroup$ – Rubertos Nov 25 '14 at 6:55
  • $\begingroup$ No. For the first lifting, you have a map $f:\{0\} \to \{e_0\}$, $f(0)=e_0$ and a map $F_0:\{0\} \times [0,1] \to X$ such that $(p\circ f)(y)=F(y,0)$ for every $y \in \{0\}$ (namely $(p\circ f)(0)=p(e_0)=F(0,0)$). This allows you to lift $F_0$ in order to obtain $G_0$. $\endgroup$ – arovai Nov 25 '14 at 17:44
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Set $Y=[0,1]$. Use path lifting to define $f$. Then apply the homotopy lifting theorem.

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