[proofplan] Uniqueness follows immediately from the [Uniqueness of Lifts theorem](/theorems/1885) applied to the connected space $I$, so the content of the proof is existence. We establish existence by a connectedness argument: the set $S \subseteq I$ of times $s$ such that a lift of $\gamma|_{[0, s]}$ with the prescribed initial value exists is shown to be nonempty, open, and closed. Nonemptiness is immediate from $s = 0$. Openness uses an evenly covered neighbourhood of $\gamma(s)$ and the sheet through the current value of the lift to extend $\tilde\gamma$ by composing $\gamma$ with the inverse homeomorphism on that sheet. Closedness also uses an evenly covered neighbourhood of $\gamma(s)$, but now we use it to produce a lift on a slightly earlier open interval and then check that the two constructions agree on their overlap — so the construction extends across $s$ from below. Connectedness of $I$ then forces $S = I$, giving a lift on all of $I$. [/proofplan]

[step:Fix notation and translate existence into a connectedness argument] Recall the [covering space](/page/Covering%20Space) setup: $p: \tilde X \to X$ is continuous and every point of $X$ has an open neighbourhood $U$ such that $p^{-1}(U) = \coprod_\alpha U_\alpha$ with each restriction $p|_{U_\alpha}: U_\alpha \to U$ a homeomorphism (the sheets over $U$). Define \begin{align*} S := \{s \in I : \text{there exists a continuous } \tilde\gamma_s: [0, s] \to \tilde X \text{ with } \tilde\gamma_s(0) = \tilde x_0 \text{ and } p \circ \tilde\gamma_s = \gamma|_{[0, s]}\}. \end{align*} By convention, at $s = 0$ the domain $[0, 0] = \{0\}$ and the map $\tilde\gamma_0(0) = \tilde x_0$ works trivially, so $0 \in S$. We aim to prove $S = I$, from which the existence of a lift on all of $I$ follows by taking $s = 1$. Uniqueness of the lift is then immediate: if $\tilde\gamma$ and $\tilde\gamma'$ are two lifts of $\gamma$ with $\tilde\gamma(0) = \tilde\gamma'(0) = \tilde x_0$, then by the [Uniqueness of Lifts theorem](/theorems/1885) applied to the continuous map $\gamma: I \to X$ and the connected space $I$, the agreement set $\{s \in I : \tilde\gamma(s) = \tilde\gamma'(s)\}$ is clopen in $I$; since $0$ is in the set, it is nonempty, and by connectedness of $I$ it is all of $I$, so $\tilde\gamma = \tilde\gamma'$. (Connectedness of $I$ is standard: $I = [0, 1]$ is a connected subset of $\mathbb{R}$.) Therefore we concentrate on existence for the rest of the proof. [/step]

[step:Show $S$ is open in $I$]Fix $s \in S$ with $s < 1$ (the case $s = 1$ is handled automatically by the claim $1 \in S$ below). Let $\tilde\gamma_s: [0, s] \to \tilde X$ be a lift as in the definition of $S$. We will extend $\tilde\gamma_s$ to a lift on $[0, s + \varepsilon]$ for some $\varepsilon > 0$, which shows the interval $[0, s + \varepsilon) \cap I$ is contained in $S$ — witnessing openness of $S$ at $s$. Choose an evenly covered open neighbourhood $U$ of $\gamma(s)$ in $X$ with $p^{-1}(U) = \coprod_\alpha U_\alpha$. Let $\tilde\gamma_s(s) \in U_\beta$ (exactly one $\beta$ by disjointness of sheets), and write \begin{align*} q := p|_{U_\beta}: U_\beta \to U, \end{align*} a homeomorphism. By continuity of $\gamma: I \to X$ at $s$, the preimage $\gamma^{-1}(U) \subseteq I$ is open and contains $s$, so it contains an open sub-interval $(s - \delta, s + \delta) \cap I$ for some $\delta > 0$. Choose any $0 < \varepsilon < \delta$ with $s + \varepsilon \le 1$, and set \begin{align*} J := [0, s + \varepsilon] \subseteq I. \end{align*} Then $\gamma(J \cap [s, s + \varepsilon]) = \gamma([s, s + \varepsilon]) \subseteq U$ by the continuity bound. Define the extension \begin{align*} \tilde\gamma_{s+\varepsilon}: J &\to \tilde X \\ t &\mapsto \begin{cases} \tilde\gamma_s(t) & t \in [0, s], \\ q^{-1}(\gamma(t)) & t \in [s, s + \varepsilon]. \end{cases} \end{align*} We verify the four conditions that this is a valid member of $S$. *Well-defined at $t = s$.* Both pieces give $q^{-1}(\gamma(s)) = \tilde\gamma_s(s)$: the top piece gives $\tilde\gamma_s(s)$ directly; the bottom piece gives $q^{-1}(\gamma(s))$, and since $q(\tilde\gamma_s(s)) = p(\tilde\gamma_s(s)) = \gamma(s)$ with $\tilde\gamma_s(s) \in U_\beta$, injectivity of $q$ gives $q^{-1}(\gamma(s)) = \tilde\gamma_s(s)$. *Continuity.* Both pieces are continuous (the first by hypothesis; the second is the composition of continuous maps $\gamma$ and $q^{-1}$), and they agree at the common point $s$ of the closed sets $[0, s]$ and $[s, s + \varepsilon]$ that cover $J$. The [pasting lemma](/theorems/???) — stating that a map defined on a finite closed cover is continuous iff its restriction to each piece is — therefore applies (the two closed sets $[0, s]$ and $[s, s + \varepsilon]$ cover $J$; both restrictions are continuous; they agree on $[0, s] \cap [s, s + \varepsilon] = \{s\}$), and the resulting map is continuous on $J$. *Initial condition.* $\tilde\gamma_{s+\varepsilon}(0) = \tilde\gamma_s(0) = \tilde x_0$ by the first piece. *Lift property.* For $t \in [0, s]$, $p(\tilde\gamma_{s+\varepsilon}(t)) = p(\tilde\gamma_s(t)) = \gamma(t)$ by hypothesis on $\tilde\gamma_s$. For $t \in [s, s + \varepsilon]$, $p(\tilde\gamma_{s+\varepsilon}(t)) = p(q^{-1}(\gamma(t))) = q(q^{-1}(\gamma(t))) = \gamma(t)$, using $p|_{U_\beta} = q$. This shows $s + \varepsilon \in S$, and by the same extension $[s, s + \varepsilon] \subseteq S$. Hence $S$ contains an open-in-$I$ neighbourhood of $s$, so $S$ is open at $s$. As $s \in S$ was arbitrary, $S$ is open in $I$.[/step]

[guided]The extension procedure is local: we use an evenly covered neighbourhood of $\gamma(s)$ and the sheet $U_\beta$ through $\tilde\gamma_s(s)$ to define the lift on $[s, s + \varepsilon]$ by inverting $p$ in that sheet. The uniqueness-of-lifts theorem (or direct injectivity of $q$) makes the two pieces match up continuously at $s$. Why is it important that $\gamma$ maps $[s, s + \varepsilon]$ entirely into $U$? Because the formula $q^{-1}(\gamma(t))$ requires $\gamma(t) \in U$ in order for $q^{-1}(\gamma(t))$ to make sense. If $\gamma(t)$ strayed outside $U$, the formula would be undefined — we would need to pass through a different evenly covered chart and continue the argument there. The continuity of $\gamma$ at $s$ is what gives us the existence of such a small interval $[s, s + \varepsilon]$ on which $\gamma$ stays in the chart. Why use the pasting lemma rather than a direct $\varepsilon$-$\delta$ argument? Because the two pieces of the definition are themselves continuous and agree at the glue point, the pasting lemma is the natural tool. The hypothesis is that $[0, s]$ and $[s, s + \varepsilon]$ are both closed in $J$ — indeed they are, as intersections of closed subsets of $\mathbb{R}$ with $J$. The glue point $\{s\}$ is the intersection of the two, and the two pieces both take the value $\tilde\gamma_s(s) = q^{-1}(\gamma(s))$ there.[/guided]

[step:Show $S$ is closed in $I$]Let $s \in \overline S \cap I$; we will show $s \in S$. Since $0 \in S$, we may assume $s > 0$. Choose an evenly covered open neighbourhood $U$ of $\gamma(s)$ in $X$, with $p^{-1}(U) = \coprod_\alpha U_\alpha$, and choose $\delta > 0$ with $\gamma((s - \delta, s + \delta) \cap I) \subseteq U$ by continuity of $\gamma$ at $s$. Since $s \in \overline S$, the interval $(s - \delta, s]$ meets $S$: pick $s_0 \in S$ with $s - \delta < s_0 \le s$. If $s_0 = s$ then $s \in S$ and we are done, so assume $s_0 < s$. Let $\tilde\gamma_{s_0}: [0, s_0] \to \tilde X$ be a lift with $\tilde\gamma_{s_0}(0) = \tilde x_0$. Let $U_\beta$ be the sheet of $p^{-1}(U)$ containing $\tilde\gamma_{s_0}(s_0)$ (exactly one such sheet by disjointness; existence follows from $\gamma(s_0) \in U$ and $p(\tilde\gamma_{s_0}(s_0)) = \gamma(s_0)$, so $\tilde\gamma_{s_0}(s_0) \in p^{-1}(U)$). Write \begin{align*} q := p|_{U_\beta}: U_\beta \to U, \end{align*} a homeomorphism. Define \begin{align*} \tilde\gamma_s: [0, s] &\to \tilde X \\ t &\mapsto \begin{cases} \tilde\gamma_{s_0}(t) & t \in [0, s_0], \\ q^{-1}(\gamma(t)) & t \in [s_0, s]. \end{cases} \end{align*} For this to be well-defined we verify that the second piece is meaningful on $[s_0, s]$: since $[s_0, s] \subseteq (s - \delta, s + \delta) \cap I$ (using $s - \delta < s_0$), we have $\gamma([s_0, s]) \subseteq U$, so $q^{-1}(\gamma(t))$ is defined for all $t \in [s_0, s]$. We verify the four conditions. *Well-defined at $t = s_0$.* The top piece gives $\tilde\gamma_{s_0}(s_0)$. The bottom piece gives $q^{-1}(\gamma(s_0))$. Since $q(\tilde\gamma_{s_0}(s_0)) = p(\tilde\gamma_{s_0}(s_0)) = \gamma(s_0)$ and $\tilde\gamma_{s_0}(s_0) \in U_\beta$, injectivity of $q$ gives $q^{-1}(\gamma(s_0)) = \tilde\gamma_{s_0}(s_0)$, so the two agree. *Continuity.* By the [pasting lemma](/theorems/???) applied to the closed cover $[0, s] = [0, s_0] \cup [s_0, s]$ with continuous restrictions agreeing on $[0, s_0] \cap [s_0, s] = \{s_0\}$. *Initial condition.* $\tilde\gamma_s(0) = \tilde\gamma_{s_0}(0) = \tilde x_0$. *Lift property.* Identical to the openness argument: both pieces of the definition satisfy $p \circ \tilde\gamma_s = \gamma$ on their respective subintervals. This shows $s \in S$. As $s \in \overline S \cap I$ was arbitrary, $\overline S \cap I \subseteq S$, so $S = \overline S \cap I$, i.e. $S$ is closed in $I$.[/step]

[guided]The closedness argument is structurally the same as the openness argument, but traversed in the reverse direction: we extend a lift from $[0, s_0]$ forward to $[0, s]$ using an evenly covered neighbourhood of $\gamma(s)$, where $s_0 < s$ is any approximating point from $S$. The crucial detail is that the **same** evenly covered chart $U$ of $\gamma(s)$ works for extending from *any* sufficiently close $s_0 \in S$. This is because once $U$ and $\delta > 0$ are fixed so that $\gamma((s - \delta, s + \delta) \cap I) \subseteq U$, any $s_0$ in this interval automatically satisfies $\gamma([s_0, s]) \subseteq U$. Why can we insist that the sheet $U_\beta$ contains $\tilde\gamma_{s_0}(s_0)$? Because $\tilde\gamma_{s_0}(s_0) \in p^{-1}(U)$ (since $\gamma(s_0) \in U$ and $p(\tilde\gamma_{s_0}(s_0)) = \gamma(s_0)$), and $p^{-1}(U) = \coprod_\alpha U_\alpha$, so $\tilde\gamma_{s_0}(s_0)$ lies in exactly one sheet. We call that sheet $U_\beta$. A potential worry: does the choice of $\tilde\gamma_{s_0}$ matter? In principle there could be several lifts of $\gamma|_{[0, s_0]}$ and we would obtain several extensions. The [Uniqueness of Lifts theorem](/theorems/1885) applied to $[0, s_0]$ (a connected interval) with the shared initial condition $\tilde x_0$ tells us all such lifts agree on $[0, s_0]$, so the choice is immaterial — but in this argument we only need existence of one such extension, so we do not need to invoke uniqueness here.[/guided]

[step:Conclude that $S = I$ by connectedness, and deduce existence and uniqueness] We have shown $S \subseteq I$ is nonempty (Step 1), open in $I$ (Step 2), and closed in $I$ (Step 3). Since the interval $I = [0, 1]$ is connected (a standard connected subset of $\mathbb{R}$), the only nonempty clopen subset of $I$ is $I$ itself. Therefore $S = I$, and in particular $1 \in S$. Thus there exists a continuous map $\tilde\gamma: I \to \tilde X$ with $\tilde\gamma(0) = \tilde x_0$ and $p \circ \tilde\gamma = \gamma$, proving existence. Uniqueness was established in Step 1 by the [Uniqueness of Lifts theorem](/theorems/1885): any two lifts starting at $\tilde x_0$ must agree on all of $I$. This completes the proof of the Path Lifting Lemma. [/step]