----- > [!proposition] Proposition. ([[paths lift uniquely under covering maps]]) > Let $p: E \to B$ be a [[covering space|covering map]] between [[topological space|topological spaces]] $E$ and $B$, let $p(e_{0}):=b_{0}$. Any [[parameterized curve]] $f:[0,1] \to B$ beginning at $b_{0}$ has a unique [[lifting]] to a [[parameterized curve]] $\tilde{f}$ in $E$ beginning at $e_{0}$. > [!NOTE] Remark. > This is a corollary of the more general result [[the homotopy lifting lemma]], but a proof 'from scratch' for this special case is provided below. > [!proof]- Proof. ([[paths lift uniquely under covering maps]]) > **Existence.** $p$ guarantees that each element of $B$ admits a [[neighborhood]] $U$ [[evenly covered]] by $p$. Cover $B$ by such open sets $U$. > Using the [[Lebesgue number lemma]], find a subdivision of $[0,1]$, say $s_{0},\dots,s_{n}$, such that for each $i$ the set $f([s_{i}, s_{i+1}])$ lies in such an open set $U$. We define the [[lifting]] $\tilde{f}:[0,1] \to E$ step by step. > First we of course define $\tilde{f}(0):=e_{0}$. Then, given $s_{i}$ and supposing $\tilde{f}$ is defined for all $0 \leq s \leq s_{i}$, we define $\tilde{f}$ on $[s_{i},s_{i+1}]$ as follows. Since $U$ is [[evenly covered]] by $p$, $p ^{-1}(U)$ equals a partition of slices $\{ V_{\alpha} \}$ s.t. $p |_{V_{\alpha}}$ is a [[homeomorphism]] onto $U$ for each $\alpha$. Choose the slice $V_{0}$ containing $\tilde{f}(s_{i})$. Now define $\tilde{f}(s):=(p |_{V_{0}})^{-1}\big( f(s) \big)\text{ for } s \in [s_{i}, s_{i+1}].$ Because $p |_{V_{0}}:V_{0} \to U$ is a [[homeomorphism]], $\tilde{f}$ will be [[continuous]] on $[s_{i},s_{i+1}]$. Continuing in this way, we define $\tilde{f}$ on all of $[0,1]$. [[continuous|Continuity]] of $\tilde{f}$ follows from [[the pasting lemma]]; the fact that $p \circ \tilde{f}=f$ is immediate from the definition of $\tilde{f}$. > **Uniqueness.** This is also proved step by step. Suppose $\tilde{\tilde{f}}$ is another [[lifting]] of $f$ beginning at $e_{0}$. Then $\tilde{\tilde{f}}(e_{0})=\tilde{f}(e_{0})$. Suppose that $\tilde{\tilde{f}}(s)=\tilde{f}(s)$ for all $s$ s.t. $0 \leq s \leq s_{i}$. Obtain $V_{0}$ as before. Since $\tilde{\tilde{f}}$ is lifting of $f$, it must carry the [[closed interval|interval]] $[s_{i},s_{i+1}]$ into the set $p ^{-1}(U)=\bigsqcup_{\alpha}V_{\alpha}$. The sets $V_{\alpha}$ are open and disjoint; [[connected subspace of separated set lies in one constituent|because]] $\tilde{\tilde{f}}([s_{i},s_{i+1}])$ is [[connected]] it must lie entirely in one of the sets $V_{\alpha}$. Because $\tilde{\tilde{f}}(s_{i})=\tilde{f}(s_{i}) \in V_{0}$, this means $\tilde{\tilde{f}}([s_{i},s_{i+1}]) \in V_{0}$. Thus, for all $s \in [s_{i},s_{i+1}]$, $\tilde{\tilde{f}}(s)=y$ for some $y \in (p |_{V_{0}})^{-1}(f(s))$. But $p |_{V_{0}}$ is a [[homeomorphism]]: there is only one such $y$, namely, $\tilde{f}(s)$. Hence $\tilde{\tilde{f}}(s)=\tilde{f}(s)$. ----- #### ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag