(Theorem) Path and homotopy lifting

Overview and statement of the theorems

Theorem: Path lifting

Let $p:X\to Y$ be a covering map, $\gamma :I\to Y$ a path and let $x\in X$ be such that $p(x)=\gamma (0)$. Then there exists precisely one path $\tilde{\gamma }:I\to X$ such that $p\circ \tilde{\gamma }=\gamma$ and $\tilde{\gamma }(0)=x$.

Theorem: Homotopy lifting

Let $p:X\to Y$ be a covering map, $W$ any topological space and $H:W\times I\to Y$ be a homotopy. Let $g:W\to X$ be such that $p\circ g=H(-,0)$. Then there exists precisely one homotopy $\tilde{H}:W\times I\to X$ with $p\circ \tilde{H}=H$ and $\tilde{H}(-,0)=g$.

Proofs from Algebraic Topology.

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Proof of path lifting

Theorem: Path lifting

Let $p:X\to Y$ be a covering map, $\gamma :I\to Y$ a path and let $x\in X$ be such that $p(x)=\gamma (0)$. Then there exists precisely one path $\tilde{\gamma }:I\to X$ such that $p\circ \tilde{\gamma }=\gamma$ and $\tilde{\gamma }(0)=x$.

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Claim 1: The path and homotopy lifting theorems hold when $p$ is a trivial covering map, i.e., the preimage $p^{-1}(Y)$ is a disjoint union of copies of $Y$.

Claim 2: The theorems hold when the image $H(W\times I)\subseteq Y$ is evenly covered.

Claim 3: Given the assumptions of the homotopy lifting theorem, let $[a,b]\subseteq I$ be subinterval such that $H(W\times [a,b])\subseteq Y$ is evenly covered, and ${\tilde{H}}_a:W\times [0,a]\to X$ be a map such that $p\circ {\tilde{H}}_a=H{|}_{W\times [0,a]}$. Then there exists precisely one map ${\tilde{H}}_b:W\times [0,b]\to X$ with both ${\tilde{H}}_b{|}_{W\times [0,a]}={\tilde{H}}_{a}p\circ {\tilde{H}}_b=H{|}_{W\times [0,b]}.$

Claim 4: Let $\mathcal{U}$ be an open cover of $[0,1]$. Then there exists $n\in \mathbb{N}$ such that for each $i=1,\dots ,n$, there is an open subset $U_i\in \mathcal{U}$ such that $[(i-1)/n,i/n]\in U_i$. In other words, $1/n$ is a Lebesgue number for the covering.

Proof.   Since $\mathcal{U}$ is a covering, for each $x\in [0,1]$ we may find ${ϵ}_x>0$ and a neighborhood $U_x\in \mathcal{U}$ such that $[0,1]\cap (x-{ϵ}_x,x+{ϵ}_x)\subseteq U_x.$ By varying $x$, we can construct another open cover of $[0,1]$ using the sets $V_x=[0,1]\cap (x-{ϵ}_x/2,x+{ϵ}_x/2)\subseteq U_x,$ and by compactness there is some $m\in \mathbb{N}$ such that ${\bigcup }_{i=1}^m{V}_{x_i}$ is a finite cover of $[0,1]$, and we can therefore choose $ϵ=\min \{{ϵ}_{x_i}:1\le j\le m\}$. Now for any $y\in [0,1]$, we have $y\in V_{x_j}$ for some $j$, so by our choice of $ϵ$ we have $(y-ϵ/2,y+ϵ/2)\subseteq (y-{ϵ}_{x_j}/2,y+{ϵ}_{x_j}/2)\subseteq V_{x_j},$ and hence $[0,1]\cap (y-ϵ/2,y+ϵ/2)\subseteq U_{x_j}$. Finally, let $n\in \mathbb{N}$ be large enough so that $1/n<ϵ/2$. Then for all $i=1,\dots ,n$, we have $[(i-1)/n,i/n]\subseteq (i/n-ϵ/2,i/n+ϵ/2),$ and the claim follows. $\square$

Proof of path lifting.   Since each point of the path $\gamma (t)\in Y$ has an evenly covered neighborhood (i.e., whose preimage is a disjoint union of open sets, each of which is mapped homeomorphically onto the base neighborhood), our goal is to find a cover of $[0,1]$ by open sets whose image in $Y$ is evenly covered. It follows from Claim 4 that we want a natural number $n$ such that $\gamma ([(i-1)/n,i/n])\subseteq Y$ is evenly covered for all $i=1,\dots ,n$.

• Uniqueness: Suppose we have a lift $\tilde{\gamma }:I\to X$ such that $\tilde{\gamma }(0)=x$ and $p\circ \tilde{\gamma }=\gamma$. We show by induction that ${\tilde{\gamma }}^{\prime }$ is another lift, then $\tilde{\gamma }{|}_{[0,i/n]}=\tilde{\gamma }’{|}_{[0,i/n]}$.
  • Base case: If $i=0$, then both maps send $0$ to $x$.
  • Inductive step: Suppose $\tilde{\gamma }{|}_{[0,(i-1)/n]}=\tilde{\gamma }’{|}_{[0,(i-1)/n]}$. Then uniqueness follows by applying Claim 3 with $a=(i-1)/n$, $b=i/n$, and taking $W$ to be the one-point space.
• Existence: We use induction to construct maps ${\tilde{\gamma }}_i:I\to X$ such that each satisfy $\tilde{\gamma }(0)=x$ and $p\circ \tilde{\gamma }=\gamma {|}_{[0,i/n]}$, at the end setting $\tilde{\gamma }={\tilde{\gamma }}_n$.
  • Base case: For $i=0$, we let ${\tilde{\gamma }}_0=x$.
  • Inductive step: Suppose ${\tilde{\gamma }}_{i-1}$ has been constructed. Then we may use the existence part of Claim 3 with $a=(i-1)/n$, $b=i/n$, and $W$ as the one-point space to construct ${\tilde{\gamma }}_i$ with precisely the desired properties. $\square$

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Proof of homotopy lifting

Theorem: Homotopy lifting

Let $p:X\to Y$ be a covering map, $W$ any topological space and $H:W\times I\to Y$ be a homotopy. Let $g:W\to X$ be such that $p\circ g=H(-,0)$. Then there exists precisely one homotopy $\tilde{H}:W\times I\to X$ with $p\circ \tilde{H}=H$ and $\tilde{H}(-,0)=g$.

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Notes

• Moreover, if $H$ is a homotopy relative to a subspace $V\subseteq W$, then so is $\tilde{H}$.
  • Suffices to prove a non-realtive version