----- Here and as usual, $I=[0,1]$. > [!proposition] Proposition. ([[path homotopies lift uniquely under covering maps]]) > Let $p: E \to B$ be a [[covering space|covering map]]; let $p(e_{0})=b_{0}$. Let the map $F: I \times I \to B$ be [[continuous]], with $F(0,0)=b_{0}$. There is a unique [[lifting]] of $F$ to a [[continuous]] map $\tilde{F}:I \times I \to E$ > such that $\tilde{F}(0,0)=e_{0}$. If $F$ is a [[path homotopy]] then so is $\tilde{F}$. > [!note] Remark. > A similar result is a corollary of [[the homotopy lifting lemma]]. Its statement and simple proof are follows. > > >[!proposition] ([[path homotopies lift uniquely under covering maps]], similar result) > Let $p: E \to B$ be a [[covering space|covering map]], $\gamma,\gamma':I \to B$ paths from $b_{0}$ to $b_{1}$, and $\tilde{\gamma}, \tilde{\gamma}': I \to X$ be [[lifting|lifts]] starting at $\tilde{b}_{0} \in p ^{-1}(b_{0})$. If $\gamma \simeq \gamma'$ as [[parameterized curve|paths]], then $\tilde{\gamma} \simeq \tilde{\gamma}'$ as paths; in particular $\tilde{\gamma}(1)=\tilde{\gamma}'(1)$. > \ > **Proof.** [[The homotopy lifting lemma]] promises a unique $\widetilde{H}: I \times I \to E$ such that $\widetilde{H}(-,0)=\tilde{\gamma}(-) \text{ and } p \circ \widetilde{H}=H.$ Now, >>- $p \circ \widetilde{H}(-,1)=H(-, 1)=\gamma'$, meaning that $H(-,1)$ is a lift of $\gamma'$ and by uniqueness we've $\widetilde{H}(-,1)=\gamma'$. >>- $p \circ \widetilde{H}(0, -)=H(0,-)$ is constant because $H(0,-)$ is; conclude $\widetilde{H}(0,-)$ is constant. Same goes for $\widetilde{H}(1,-)$. ^proposition ^note > [!proof]- Proof. ([[path homotopies lift uniquely under covering maps]]) > **Existence.** Given $F$, we first define $\tilde{F}(0,0):=e_{0}$. Next, we use [[paths lift uniquely under covering maps]] to extend $\tilde{F}$ to the left-hand edge $\{ 0 \} \times I$ and the bottom edge $I \times \{ 0 \}$ of $I \times I$. > Using the [[Lebesgue number lemma]], choose [[partition|subdivisions]] $\begin{align} s_{0} < s_{1} < \dots < s_{m} \\ t_{0} < t_{1} < \dots < t_{n} \end{align}$ of $I$ fine enough that each [[rectangle]] $I_{i} \times J_{j} = [s_{i-1}, s_{i}] \times [t_{j-1}, t_{j}]$ we have $F(I_{i} \times J_{j}) \subset U$ for some open subset $U$ of $B$ [[evenly covered]] by $p$. > In general, given $i_{0}$ and $j_{0}$, assume that $\tilde{F}$ is defined on the set $A$ which is the union of $\{ 0 \} \times I$ and $I \times \{ 0 \}$ and all the rectangles 'previous' to $I_{i_{0}} \times J_{j_{0}}$ (those rectangles $I_{i} \times J_{j}$ for which $j<j_{0}$ and those for which $j=j_{0}$ and $i<i_{0}$). Assume also that $\tilde{F}$ is a [[continuous]] [[lifting]] of $F |_{A}$. We define $\tilde{F}$ on $I_{i_{0}} \times J_{j_{0}}$. Choose an open set $U$ of $B$ containing $F(I_{i_{0}} \times J_{j_{0}})$ that is [[evenly covered]] by $p$. Partition $p ^{-1}(U)$ [[evenly covered|into slices]] $\{ V_{\alpha} \}$, where $p |_{V_{\alpha}}$ is a [[homeomorphism]] of $V_{\alpha}$ onto $U$. Now $\tilde{F}$ is already defined on the set $C=A \cap (I_{i_{0}} \times J_{j_{0}})$. This set is the union of the left and bottom edges of the [[rectangle]], [[arbitrary union of nontrivially-intersecting connected subspaces is connected|so]] it is [[connected]]. Therefore $\tilde{F}(C)$ [[continuity preserves connectedness|is connected]] and so [[connected subspace of separated set lies in one constituent|lies entirely]] in one of the slices, say, $\tilde{F}(C) \subset V_{0}$. > ![[CleanShot 2024-03-30 at [email protected]]] > Let $p_{0}:V_{0} \to U$ denote the restriction of $p$ to $V_{0}$. Since $\tilde{F}$ is 'already' a lifting of $F |_{A}$, we know that for $x \in C$, $p_{0}(\tilde{F}(x)) = F(x)$ so that $\tilde{F}(x)=p_{0}^{-1}(F(x))$. Hence we may safely extend $\tilde{F}$ by defining $\tilde{F}(x)=p_{0}^{-1}(F(x))$ for $x \in I_{i_{0}} \times J_{j_{0}}$. The extended map will be [[connected]] by [[the pasting lemma]]. Continuing in this way, we define $\tilde{F}$ on all of $I^{2}.$ > **Uniqueness.** To check uniqueness, note that at each step of construction $\tilde{F}$, as we extend $\tilde{F}$ from $(0,0)$ first to the bottom and left edges of $I^{2}$ and then to the rectangles $I_{i} \times J_{j}$ one by one, there is only one way to extend $\tilde{F}$ continuously. > **Path homotopy preservation.** Suppose additionally that $F$ is a [[path homotopy]]. WTS $\tilde{F}$ is too. Since $F$ is a [[path homotopy]], we have $F(\{ 0 \} \times I)=b_{0} \in B$. Now, $\tilde{F}(\{ 0 \} \times I)=p_{0}^{-1}(b_{0})$ [[covering space#^0cf198|but this set has]] the [[discrete topology]] as a [[subspace topology|subspace]] of $E$. Since $\{ 0 \} \times I$ is [[connected]] and $\tilde{F}$ is [[continuous]], $\tilde{F}(\{ 0 \} \times I)$ is [[connected]] and thus must be a singleton. Likewise $\tilde{F}(\{ 1 \} \times I)$ must be a singleton. Thus $\tilde{F}$ is a [[path homotopy]]. ----- #### scratch: This is a corollary of [[the homotopy lifting lemma]]. Indeed, let us say $\gamma$ and $\gamma'$ are paths in $B$, [[homotopy|homotopic]] via $H: I \times I \to B$. Denote $\gamma(0)=\gamma'(0)=x_{0}$ and choose $\tilde{x}_{0} \in E$ satisfying $p(\tilde{x}_{0})=x_{0}$. Then the 'path-lifting' corollary in [[the homotopy lifting lemma]] promises unique [[parameterized curve|paths]] $\tilde{\gamma},\tilde{\gamma}':I \to E$ such that - $\tilde{\gamma}(0)=\tilde{\gamma}'(0)=\tilde{x}_{0}$ - $p \circ \tilde{\gamma}=\gamma$ and $p \circ \tilde{\gamma}'=\gamma'$ (i.e., [[lifting|lifts]]) ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```