Theorem (Path Lifting): Let $p : \tilde{X} \to X$ be a covering space, $\gamma : [0,1] \to X$ a path with $\gamma(0) = x_0$, and $\tilde{x}_0 \in p^{-1}(x_0)$. Then there exists a unique path $\tilde{\gamma} : [0,1] \to \tilde{X}$ with $p \circ \tilde{\gamma} = \gamma$ and $\tilde{\gamma}(0) = \tilde{x}_0$.

Existence: We construct the lift using the Lebesgue number lemma. Since $[0,1]$ is compact, $\gamma^{-1}$ of the cover of $\gamma([0,1])$ by evenly covered neighborhoods has a Lebesgue number $\delta > 0$.

Partition $[0,1]$ into subintervals $I_k = [t_{k-1}, t_k]$ with $|I_k| < \delta$, so $\gamma(I_k)$ lies in an evenly covered neighborhood $U_k$.

Inductive construction:

• Base: Start with $\tilde{\gamma}(0) = \tilde{x}_0$.
• Step: Suppose $\tilde{\gamma}$ is defined on $[0, t_k]$. The set $p^{-1}(U_{k+1})$ is a disjoint union of open sets $\tilde{U}_\alpha$, each mapped homeomorphically onto $U_{k+1}$ by $p$.
• Since $\tilde{\gamma}(t_k) \in p^{-1}(\gamma(t_k)) \cap p^{-1}(U_{k+1})$, it lies in exactly one sheet $\tilde{U}_\alpha$.
• Extend $\tilde{\gamma}$ over $I_{k+1}$ using $(p|_{\tilde{U}_\alpha})^{-1} \circ \gamma$ on $I_{k+1}$.

By induction, $\tilde{\gamma}$ is defined on all of $[0,1]$, and by construction $p \circ \tilde{\gamma} = \gamma$.

Uniqueness: Suppose $\tilde{\gamma}_1, \tilde{\gamma}_2$ are two lifts starting at $\tilde{x}_0$. Let $A = \{t \in [0,1] : \tilde{\gamma}_1(t) = \tilde{\gamma}_2(t)\}$.

We show $A$ is both open and closed in $[0,1]$:

• Closed: If $t_n \in A$ with $t_n \to t$, then by continuity $\tilde{\gamma}_1(t) = \lim \tilde{\gamma}_1(t_n) = \lim \tilde{\gamma}_2(t_n) = \tilde{\gamma}_2(t)$, so $t \in A$.
• Open: If $t \in A$, let $U$ be an evenly covered neighborhood of $\gamma(t)$. Then $\tilde{\gamma}_1(t) = \tilde{\gamma}_2(t)$ lies in a unique sheet $\tilde{U}$ of $p^{-1}(U)$. For small $\epsilon$, both $\tilde{\gamma}_1$ and $\tilde{\gamma}_2$ map $[t-\epsilon, t+\epsilon]$ into $\tilde{U}$. Since $p$ is injective on $\tilde{U}$ and $p \circ \tilde{\gamma}_1 = p \circ \tilde{\gamma}_2 = \gamma$, we have $\tilde{\gamma}_1 = \tilde{\gamma}_2$ on $[t-\epsilon, t+\epsilon]$. Thus $A$ is open.

Since $A$ is non-empty (contains 0), open, and closed in the connected space $[0,1]$, we have $A = [0,1]$. Therefore $\tilde{\gamma}_1 = \tilde{\gamma}_2$. ∎