455: Lifts, That Start at Same Point, of Path-Homotopic Paths Are Path-Homotopic

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A description/proof of that lifts, that start at same point, of path-homotopic paths are path-homotopic

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of covering map.
• The reader knows a definition of lift of continuous map with respect to covering map.
• The reader admits the proposition that for any covering map, there is the unique lift of any continuous map from the finite product of any closed real intervals for each initial value.

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Target Context

• The reader will have a description and a proof of the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any connected and locally path-connected topological spaces, $T_1,T_2$, any covering map, $\pi :T_1\to T_2$, which means that $\pi$ is continuous and surjective and around any point, $p\in T_2$, there is a neighborhood, $N_p\subseteq T_2$, that is evenly covered by $\pi$, any closed interval, $T_3:=[r_1,r_2]$, any path-homotopic paths, $f_1:T_3\to T_2$ and $f_2:T_3\to T_2$, and any point $p_0\in T_1$ such that $\pi (p_0)=f_1(r_1)=f_2(r_1)$, the lifts, that start at $p_0$, of $f_1$ and $f_2$, $f_1^{\prime }:T_3\to T_1$ and $f_2^{\prime }:T_3\to T_1$, are path-homotopic.

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2: Proof

There is a homotopy relative to $\{r_1,r_2\}$, $H:T_3\times I\to T_2$, $H(p,0)=f_1(p)$, $H(p,1)=f_2(p)$, $H(r_1,s)=f_1(r_1)=f_2(r_1)$, $H(r_2,s)=f_1(r_2)=f_2(r_2)$.

$f_1^{\prime }(r_1)=f_2^{\prime }(r_1)=p_0$. $\pi \circ f_i^{\prime }=f_i$.

There is the unique lift, $H^{\prime }:T_3\times I\to T_1$, of $H$ for the initial value, $H^{\prime }(r_1,0)=p_0$, by the proposition that for any covering map, there is the unique lift of any continuous map from the finite product of any closed real intervals for each initial value.

For any fixed $s\in I$, $H^{\prime }(p,s)$ is the unique lift of $H(p,s)$ for the initial value, $H^{\prime }(r_1,s)$, because $\pi \circ H^{\prime }(p,s)=H(p,s)$. For any fixed $p\in T_3$, $H^{\prime }(p,s)$ is the unique lift of $H(p,s)$ for the initial value, $H^{\prime }(p,0)$, because $\pi \circ H^{\prime }(p,s)=H(p,s)$.

$\pi \circ H^{\prime }(p,0)=H(p,0)=f_1(p)$, so, as the unique lift of $f_1$ for the initial value, $H^{\prime }(r_1,0)=p_0$, $H^{\prime }(p,0)=f_1^{\prime }(p)$. As the constant map, $s\mapsto p_0$, is a lift of $H(r_1,s)$ for the initial value, $0\mapsto p_0$, and $H^{\prime }(r_1,s)$ is the unique lift for the initial value, $H^{\prime }(r_1,0)=p_0$, $H^{\prime }(r_1,s)=p_0=f_1^{\prime }(r_1)=f_2^{\prime }(r_1)$. $\pi \circ H^{\prime }(p,1)=H(p,1)=f_2(p)$, so, as the unique lift of $f_2$ for the initial value, $H^{\prime }(r_1,1)=p_0$, $H^{\prime }(p,1)=f_2^{\prime }(p)$. As the constant map, $s\mapsto f_1^{\prime }(r_2)$, is a lift of $H(r_2,s)$ for the initial value, $0\mapsto f_1^{\prime }(r_2)$, and $H^{\prime }(r_2,s)$ is the unique lift for the initial value, $H^{\prime }(r_2,0)=f_1^{\prime }(r_2)$, $H^{\prime }(r_2,s)=f_1^{\prime }(r_2)=f_2^{\prime }(r_2)$.

So, $H^{\prime }$ is a homotopy relative to $\{r_1,r_2\}$.

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References

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