454: For Covering Map, There Is Unique Lift of Path for Each Point in Covering Map Preimage of Path Image of Point on Path Domain

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A description/proof of that for covering map, there is unique lift of path for each point in covering map preimage of path image of point on path domain

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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 in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
• The reader admits the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
• The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain.

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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, $T_3:=[r_1,r_2]$, and any path, $f:T_3\to T_2$, for any point, $p_0\in T_3$, and each point $p_0^{\prime }\in {\pi }^{-1}(f(p_0))$, there is the unique lift of $f$, $f^{\prime }$, such that $f^{\prime }(p_0)=p_0^{\prime }$.

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2: Proof

The subspace, ${\pi }^{-1}(N_p)$, may consist of multiple connected components, each denoted as ${{\pi }^{-1}(N_p)}_{\alpha }$ where $\alpha \in A_p$ where $A_p$ is a possibly uncountable indices set.

We can take an open neighborhood, $U_p\subseteq T_2$, as $N_p$, because if $N_p$ is not open, there is an open neighborhood, $U_p\subseteq N_p$, which is homeomorphic to each ${{\pi }^{-1}(U_p)}_{\alpha }$ by ${\pi }_{p,\alpha }:=\pi {|}_{{{\pi }^{-1}(U_p)}_{\alpha }}:{{\pi }^{-1}(U_p)}_{\alpha }\to U_p$, because ${\pi }_{p,\alpha }$ is obviously bijective, is continuous with the domain and the codomain regarded as the subspaces of ${{\pi }^{-1}(N_p)}_{\alpha }$ and $N_p$ respectively as a restriction of continuous $\pi {|}_{{{\pi }^{-1}(N_p)}_{\alpha }}:{{\pi }^{-1}(N_p)}_{\alpha }\to N_p$, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, and its inverse is continuous with the domain and the codomain regarded likewise as a restriction of continuous ${\pi {|}_{{{\pi }^{-1}(N_p)}_{\alpha }}}^{-1}$, likewise, but by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, those maps are continuous also with the domain and the codomain regarded as subspaces of $T_1$ and $T_2$ respectively.

For any point, $p\in T_3$, there are an $U_{f(p)}\subseteq T_2$ and $\{{{\pi }^{-1}(U_{f(p)})}_{\alpha }\subseteq T_1\}$. As $f$ is continuous, there is an open neighborhood, $U_p\subseteq T_3$, such that $f(U_p)\subseteq U_{f(p)}$, but in fact, let us take $U_p$ as an open ball, which is obviously possible. For each $\alpha$, there is the map, $f_{p,\alpha }^{\prime }:U_p\to {{\pi }^{-1}(U_{f(p)})}_{\alpha }$, $p^{\prime }\mapsto f(p^{\prime })\mapsto {{\pi }_{p,\alpha }}^{-1}(f(p^{\prime }))$ where ${{\pi }_{p,\alpha }}^{-1}:U_{f(p)}\to {{\pi }^{-1}(U_{f(p)})}_{\alpha }=(\pi {|}_{{{\pi }^{-1}(U_{f(p)})}_{\alpha }}{)}^{-1}$, continuous. So, $f_{p,\alpha }^{\prime }$ is continuous. $\{U_p\}$ covers $T_3$ and $T_3$ is compact, so, there is a finite subcover, $\{U_{p_i}\}$.

$p_0\in U_{p_1}$ without loss of generality. Let us choose $f_{p_1,{\alpha }_1}^{\prime }$ as $p_0^{\prime }\in {{\pi }^{-1}(U_{f(p_1)})}_{{\alpha }_1}$. There is a $U_{p_2}$ such that $U_{p_1}\cap U_{p_2}\ne \mathrm{\varnothing }$, without loss of generality, because $T_3$ is connected. There is a point, $p^{\prime }\in U_{p_1}\cap U_{p_2}$. Let us choose $f_{p_2,{\alpha }_2}^{\prime }$ as $f_{p_2,{\alpha }_2}^{\prime }(p^{\prime })=f_{p_1,{\alpha }_1}^{\prime }(p^{\prime })$. $U_{p_1}\cap U_{p_2}$ is connected, because $U_{p_i}$ is an open ball. $f_{p_i,{\alpha }_i}^{\prime }{|}_{U_{p_1}\cap U_{p_2}}$ is a lift of $f{|}_{U_{p_1}\cap U_{p_2}}$, and $f_{p_1,{\alpha }_1}^{\prime }{|}_{U_{p_1}\cap U_{p_2}}=f_{p_2,{\alpha }_2}^{\prime }{|}_{U_{p_1}\cap U_{p_2}}$, by the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree. Let us define $f_{p_1,p_2}^{\prime }:U_{p_1}\cup U_{p_2}\to T_1$ as $f_{p_1,p_2}^{\prime }{|}_{U_{p_i}}=f_{p_i,{\alpha }_i}^{\prime }$, continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.

There is a $U_{p_3}$, $(U_{p_1}\cup U_{p_2})\cap U_{p_3}\ne \mathrm{\varnothing }$. $p^{\prime }\in (U_{p_1}\cup U_{p_2})\cap U_{p_3}$. $p^{\prime }\in U_{p_1}$ or $p^{\prime }\in U_{p_2}$ (or both), and let us take the least $i$ such that $p^{\prime }\in U_{p_i}$. Let us choose $f_{p_3,{\alpha }_3}^{\prime }$ as $f_{p_3,{\alpha }_3}^{\prime }(p^{\prime })=f_{p_i,{\alpha }_i}^{\prime }(p^{\prime })$. $(U_{p_1}\cup U_{p_2})\cap U_{p_3}$ is connected, because $U_{p_i}$ is an open ball. Each of $f_{p_3,{\alpha }_3}^{\prime }{|}_{(U_{p_1}\cup U_{p_2})\cap U_{p_3}}$ and $f_{p_1,p_2}^{\prime }{|}_{(U_{p_1}\cup U_{p_2})\cap U_{p_3}}$ is a lift of $f{|}_{(U_{p_1}\cup U_{p_2})\cap U_{p_3}}$, and $f_{p_3,{\alpha }_3}^{\prime }{|}_{(U_{p_1}\cup U_{p_2})\cap U_{p_3}}=f_{p_1,p_2}^{\prime }{|}_{(U_{p_1}\cup U_{p_2})\cap U_{p_3}}$, by the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree. Let us define $f_{p_1,p_2,p_3}^{\prime }:U_{p_1}\cup U_{p_2}\cup U_{p_3}\to T_1$ as $f_{p_1,p_2,p_3}^{\prime }{|}_{U_{p_1}\cup U_{p_2}}=f_{p_1,p_2}^{\prime }$ and $f_{p_1,p_2,p_3}^{\prime }{|}_{U_{p_3}}=f_{p_3,{\alpha }_3}^{\prime }$, continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.

By mathematical induction, there is a lift, $f_{p_1,p_2,...,p_n}^{\prime }:U_{p_1}\cup U_{p_2}\cup ...\cup U_{p_n}=T_3\to T_1$ of, $f$.

It is the unique lift of $f$ for $p_0^{\prime }$, by the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree.

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References

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