450: For Covering Map, There Is Unique Lift of Continuous Map from Finite Product of Closed Real Intervals for Each Initial Value

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A description/proof of that for covering map, there is unique lift of continuous map from finite product of closed real intervals for each initial value

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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 Lebesgue number lemma.
• The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
• 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 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 the product of any finite number of compact topological spaces is compact.
• The reader admits the proposition that for any covering map, there is the unique lift of any continuous map from any real closed interval for each initial value.

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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 continuous map from the finite product of any closed real intervals for each initial value.

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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 $\pi$, the finite product of any some closed real intervals, $T_3=T_1^{\prime }\times T_2^{\prime }\times ...\times T_n^{\prime }$ where $T_i^{\prime }:=[r_{i,1},r_{i,2}]\subseteq \mathbb{R}$, and any continuous map, $f:T_3\to T_2$, there is the unique lift of $f$, $f^{\prime }:T_3\to T_1$, for any initial value, $p_0=f^{\prime }(r_{1,1},r_{2,1},...,r_{n,1})$ where $(r_{1,1},r_{2,1},...,r_{n,1})$ is called the origin.

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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 a $U_{f(p)}\subseteq T_2$ 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 $f(U_p)\subseteq U_{f(p)}$. For each $\alpha$, there is the map, ${{\pi }_{p,\alpha }}^{-1}:U_{f(p)}\to {{\pi }^{-1}(U_{f(p)})}_{\alpha }=(\pi {|}_{{{\pi }^{-1}(U_{f(p)})}_{\alpha }}{)}^{-1}$, continuous. $\{U_p\}$ covers $T_3$, and $T_3$ is compact, by the proposition that the product of any finite number of compact topological spaces is compact. There is a real number, $\delta$, such that any subset whose diameter is less than $\delta$ is contained in an open subset in the open cover, by the Lebesgue number lemma. Let $T_3$ be tiled starting from the origin with closed cubes whose edge length is ${\delta }^{\prime }$ that makes the diameter of each cube less than $\delta$; in fact, there can be some residues (as $r_{i,2}-r_{i,1}$ may not be any natural number multiple of ${\delta }^{\prime }$), which are tiled with smaller closed rectangular parallelepipeds whose diameters are still less than $\delta$. Then, each closed rectangular parallelepiped (including cubes) is contained in a $U_p$. Each of such closed rectangular parallelepipeds is denoted as $u_{i_1,i_2,...,i_n}$, numbered starting from the origin and $0\le i_j\le m_j$.

Let us suppose that $p_0=f^{\prime }(r_{1,1},r_{2,1},...,r_{n,1})$ is fixed.

For each $i$, there is the unique lift, $f^{\prime }{|}_{\{r_{1,1}\}\times \{r_{2,1}\}\times ...\times \{r_{i-1,1}\}\times T_i^{\prime }\times \{r_{i+1,1}\}\times ...\times \{r_{n,1}\}}(p_1,p_2,...,p_i,...,p_n)$ of $f(r_{1,1},r_{2,1},...,p_i,...,r_{n,1})$ where $r_{j,1}$ fixed, by the proposition that for any covering map, there is the unique lift of any continuous map from any real closed interval for each initial value. $S_0:={\cup }_i\{r_{1,1}\}\times \{r_{2,1}\}\times ...\times \{r_{i-1,1}\}\times T_i^{\prime }\times \{r_{i+1,1}\}\times ...\times \{r_{n,1}\}$ is the area on which the values of $f^{\prime }$ has been determined so far.

Let us define $S_{0,0,...,0}:=S_0\cup u_{0,0,...,0}$; $S_{i_1,i_2,...,i_n}:=S_{i_1,i_2,...,i_n-1}\cup u_{i_1,i_2,...,i_n}$ if $0<i_n$; $S_{i_1,i_2,...,i_{n-1},0}:=S_{i_1,i_2,...,i_{n-1}-1,m_n}\cup u_{i_1,i_2,...,i_n}$ if $0<i_{n-1}$; $S_{i_1,i_2,...,i_{n-2},0,0}:=S_{i_1,i_2,...,i_{n-2}-1,m_{n-1},m_n}\cup u_{i_1,i_2,...,i_n}$ if $0<i_{n-2}$; etc.: they are inductively defined in the order: $S_{0,0,...,0},S_{0,0,...,0,1},...,S_{0,0,...,0,m_n},S_{0,0,...,0,1,0},S_{0,0,...,0,1,1},...,S_{0,0,...,0,1,m_n},S_{0,0,...,0,2,0},S_{0,0,...,0,2,1},...,S_{0,0,...,0,2,m_n},...,S_{0,0,...,0,m_{n-1},m_n},...,S_{m_1,m_2,...,m_{n-n},m_{n-1},m_n}$.

Let us define the intersection subspaces, $S_{0,0,...,0}^{\prime }:=S_0\cap u_{0,0,...,0}$; $S_{i_1,i_2,...,i_n}^{\prime }:=S_{i_1,i_2,...,i_n-1}\cap u_{i_1,i_2,...,i_n}$ if $0<i_n$; $S_{i_1,i_2,...,i_{n-1},0}^{\prime }:=S_{i_1,i_2,...,i_{n-1}-1,m_n}\cap u_{i_1,i_2,...,i_n}$ if $0<i_{n-1}$; $S_{i_1,i_2,...,i_{n-2},0,0}:=S_{i_1,i_2,...,i_{n-2}-1,m_{n-1},m_n}\cap u_{i_1,i_2,...,i_n}$ if $0<i_{n-2}$; etc.: the intersections are taken instead of the unions in the definitions in the previous paragraph.

Note that we may hereafter show only the $0<i_n$ version when the meaning is obvious that the other cases are likewise.

Let us see that each $S_{i_1,i_2,...,i_n}^{\prime }$ is path-connected with the left-bottom point of $u_{i_1,i_2,...,i_n}$, so, it is connected. $S_{i_1,i_2,...,i_n}^{\prime }$ consists of the intersection of $u_{i_1,i_2,...,i_n}$ with $S_0$ and the intersections of $u_{i_1,i_2,...,i_n}$ with some $u_{i_1^{\prime },i_2^{\prime },...,i_n^{\prime }}$ s where $i_j-1\le i_j^{\prime }\le i_j+1$. Let us think point on $T_3$ with the indices expression: the points on $u_{i_1,i_2,...,i_n}$ are $(i_1\sim i_1+1,i_2\sim i_2+1,...,i_n\sim i_n+1)$. Each part of the intersection of $u_{i_1,i_2,...,i_n}$ with $S_0$ is the intersection of $u_{i_1,i_2,...,i_n}$ with a $\{r_{1,1}\}\times \{r_{2,1}\}\times ...\times \{r_{j-1,1}\}\times T_j^{\prime }\times \{r_{j+1,1}\}\times ...\times \{r_{n,1}\}$, which is $(0,...,0,i_j\sim i_j+1,0,...,0)$ in the indices expression (it has to be that $i_0=...=i_{j-1}=i_{j+1}=...=i_n=0$), which is path-connected with the left-bottom point, $(i_1,i_2,...,i_n)=(0,...,0,i_j,0,...,0)$. As for the intersection of $u_{i_1,i_2,...,i_n}$ with a $u_{i_1^{\prime },i_2^{\prime },...,i_n^{\prime }}$ where $i_j-1\le i_j^{\prime }\le i_j+1$, when $i_j-1\le i_j^{\prime }\le i_j$ for each $j$, the intersection is like $(i_1\sim i_1+1,i_2,...,i_n\sim i_n+1)$ where "$i_j\sim i_j+1$" like for $j=1$ is because $i_j^{\prime }=i_j$ and "$i_j$" like for $j=2$ because $i_j-1=i_j^{\prime }$, and the intersection is path-connected with the left-bottom point, $(i_1,i_2,...,i_n)$; when $i_j^{\prime }=i_j+1$ for some $j$ s, the intersection is like $(i_1\sim i_1+1,i_2,...,i_j+1,...,i_n\sim i_n+1)$, which, in fact, is not path-connected with the left-bottom point, $(i_1,i_2,...,i_n)$, but $u_{i_1,i_2,...,i_n}$ intersects also with the corresponding $u_{i_1^{\prime },i_2^{\prime },...,i_j^{\prime }-1=i_j,...,i_n}$ with all the $i_j^{\prime }$ s such that $i_j^{\prime }=i_j+1$ decremented by $1$, and the intersection is $(i_1\sim i_1+1,i_2,...,i_j\sim i_j+1,...,i_n\sim i_n+1)$, which contains the former intersection and is path-connected with the left-bottom point, $(i_1,i_2,...,i_n)$. So, each part of $S_{i_1,i_2,...,i_n}^{\prime }$ is path-connected with the left-bottom point of $u_{i_1,i_2,...,i_n}$, and $S_{i_1,i_2,...,i_n}^{\prime }$ is path-connected with the left-bottom point of $u_{i_1,i_2,...,i_n}$ as a finite union of such parts. For example, $S_{0,0,...,0}^{\prime }$ is the edges of $u_{0,0,...,0}$ that connect with the left-bottom point of $u_{0,0,...,0}$; $S_{1,0,0}^{\prime }$ is the union of the edge $(1\sim 2,0,0)$ and $(1,0\sim 1,0\sim 1)$.

$f^{\prime }{|}_{S_0}$ has been already determined. $u_{0,0,...,0}\subseteq U_p$. $f(u_{0,0,...,0})\subseteq U_{f(p)}$. As $S_{0,0,...,0}^{\prime }\subseteq S_0$, $f^{\prime }{|}_{S_0}$ has been already determined there. While $f^{\prime }{|}_{S_0}(S_{0,0,...,0}^{\prime })\subseteq {\pi }^{-1}(U_{f(p)})$, because $S_{0,0,...,0}^{\prime }\subseteq u_{0,0,...,0}$ and $\pi \circ f^{\prime }{|}_{S_0}(S_{0,0,...,0}^{\prime })=f(S_{0,0,...,0}^{\prime })\subseteq U_{f(p)}$, let us choose ${{\pi }^{-1}(U_{f(p)})}_{\alpha }$ as $f^{\prime }{|}_{S_0}(S_{0,0,...,0}^{\prime })\subseteq {{\pi }^{-1}(U_{f(p)})}_{\alpha }$, which is valid, because as $S_{0,0,...,0}^{\prime }$ is connected, $f^{\prime }{|}_{S_0}(S_{0,0,...,0}^{\prime })$ is connected. ${\pi }_{p,\alpha }:=\pi {|}_{{\pi }^{-1}(U_{f(p)}{)}_{\alpha }}$. $f^{\prime }{|}_{u_{0,0,...,0}}:=({\pi }_{p,\alpha }{)}^{-1}\circ f{|}_{u_{0,0,...,0}}$, continuous. $f^{\prime }{|}_{S_0}{|}_{S_{0,0,...,0}^{\prime }}=f^{\prime }{|}_{u_{0,0,...,0}}{|}_{S_{0,0,...,0}^{\prime }}$, because $\pi \circ f^{\prime }{|}_{S_0}{|}_{S_{0,0,...,0}^{\prime }}=f{|}_{S_{0,0,...,0}^{\prime }}=\pi \circ f^{\prime }{|}_{u_{0,0,...,0}}{|}_{S_{0,0,...,0}^{\prime }}$ but we have $f^{\prime }{|}_{S_0}{|}_{S_{0,0,...,0}^{\prime }}(S_{0,0,...,0}^{\prime }),f^{\prime }{|}_{u_{0,0,...,0}}{|}_{S_{0,0,...,0}^{\prime }}(S_{0,0,...,0}^{\prime })\subseteq {\pi }^{-1}(U_{f(p)}{)}_{\alpha }$ and ${\pi }_{p,\alpha }$ is injective. Let us define $f^{\prime }{|}_{S_{0,0,...,0}}$ as $f^{\prime }{|}_{S_{0,0,...,0}}{|}_{S_0}=f^{\prime }{|}_{S_0}$ and $f^{\prime }{|}_{S_{0,0,...,0}}{|}_{u_{0,0,...,0}}=f^{\prime }{|}_{u_{0,0,...,0}}$. So, $f^{\prime }$ is determined on $S_{0,0,...,0}$.

Let us suppose that $f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}$ has been already determined. $u_{i_1,i_2,...,i_n}\subseteq U_p$. $f(u_{i_1,i_2,...,i_n})\subseteq U_{f(p)}$. As $S_{i_1,i_2,...,i_n}^{\prime }\subseteq S_{i_1,i_2,...,i_n-1}$, $f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}$ has been already determined there. While $f^{\prime }{|}_{S_0}(S_{i_1,i_2,...,i_n}^{\prime })\subseteq {\pi }^{-1}(U_{f(p)})$, because $S_{i_1,i_2,...,i_n}^{\prime }\subseteq u_{i_1,i_2,...,i_n}$ and $\pi \circ f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}(S_{i_1,i_2,...,i_n}^{\prime })=f(S_{i_1,i_2,...,i_n}^{\prime })\subseteq U_{f(p)}$, let us choose ${{\pi }^{-1}(U_{f(p)})}_{\alpha }$ as $f^{\prime }(S_{i_1,i_2,...,i_n}^{\prime })\subseteq {{\pi }^{-1}(U_{f(p)})}_{\alpha }$, which is valid, because as $S_{i_1,i_2,...,i_n}^{\prime }$ is connected, $f^{\prime }(S_{i_1,i_2,...,i_n}^{\prime })$ is connected. ${\pi }_{p,\alpha }:=\pi {|}_{{\pi }^{-1}(U_{f(p)}{)}_{\alpha }}$. $f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}:=({\pi }_{p,\alpha }{)}^{-1}\circ f{|}_{u_{i_1,i_2,...,i_n}}$, continuous. $f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}{|}_{S_{i_1,i_2,...,i_n}^{\prime }}=f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}{|}_{S_{i_1,i_2,...,i_n}^{\prime }}$, because $\pi \circ f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}{|}_{S_{i_1,i_2,...,i_n}^{\prime }}=f{|}_{S_{i_1,i_2,...,i_n}^{\prime }}=\pi \circ f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}{|}_{S_{i_1,i_2,...,i_n}^{\prime }}$ but we have $f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}{|}_{S_{i_1,i_2,...,i_n}^{\prime }}(S_{i_1,i_2,...,i_n}^{\prime }),f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}{|}_{S_{i_1,i_2,...,i_n}^{\prime }}(S_{i_1,i_2,...,i_n}^{\prime })\subseteq {\pi }^{-1}(U_{f(p)}{)}_{\alpha }$ and ${\pi }_{p,\alpha }$ is injective. Let us define $f^{\prime }{|}_{S_{i_1,i_2,...,i_n}}$ as $f^{\prime }{|}_{S_{i_1,i_2,...,i_n}}{|}_{S_{i_1,i_2,...,i_n-1}}=f^{\prime }{|}_{S_{i_1,i_2,...,i_n-1}}$ and $f^{\prime }{|}_{S_{i_1,i_2,...,i_n}}{|}_{u_{i_1,i_2,...,i_n}}=f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}$. So, the whole $f^{\prime }$ is determined.

Let us confirm that $f^{\prime }$ is a lift of $f$. $f^{\prime }$ is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous. For any $p\in u_{i_1,i_2,...,i_n}$, $\pi \circ f^{\prime }(p)=\pi \circ ({\pi }_{p,\alpha }{)}^{-1}\circ f(p)=f(p)$.

$f^{\prime }$ is unique, because it is uniquely determined on $S_0$ and each extension to $S_{i_1,i_2,...,i_n}$ has no leeway: ${{\pi }^{-1}(U_{f(p)})}_{\alpha }$ is uniquely determined and $f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}:=({\pi }_{p,\alpha }{)}^{-1}\circ f{|}_{u_{i_1,i_2,...,i_n}}$ has to be defined so in order to satisfy $\pi \circ f^{\prime }{|}_{u_{i_1,i_2,...,i_n}}:=f{|}_{u_{i_1,i_2,...,i_n}}$.

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References

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