[proofplan] We prove the result locally along the base path and then concatenate the local solutions. On a subinterval where $\gamma$ lies in a trivializing [open set](/page/Open%20Set), a lift can be expressed uniquely by choosing a smooth local section and a smooth group-valued curve in the structure group. The horizontality condition becomes a first-order ordinary differential equation on $G$, whose local existence and uniqueness give a unique horizontal lift on that subinterval. Compactness of $[0,1]$ gives finitely many such subintervals, and uniqueness of the initial value problem makes the locally constructed lifts agree at the endpoints. [/proofplan]

[step:Choose finitely many trivializing intervals along the path] For each $t\in[0,1]$, choose an open neighbourhood $U_t\subset M$ of $\gamma(t)$ over which $P$ admits a smooth local section \begin{align*} s_t:U_t\to P. \end{align*} Since $\gamma$ is continuous, the sets $\gamma^{-1}(U_t)\subset[0,1]$ form an open cover of $[0,1]$. By the [Lebesgue Number Lemma](/theorems/952) for the compact [metric space](/page/Metric%20Space) $[0,1]$, applied to this open cover, there is a number $\delta>0$ such that every subset of $[0,1]$ with diameter less than $\delta$ is contained in some member of this cover. Choose a subdivision \begin{align*} 0=t_0<t_1<\cdots<t_N=1 \end{align*} with $t_j-t_{j-1}<\delta$ for every $j\in\{1,\dots,N\}$. Since the closed interval $[t_{j-1},t_j]$ has diameter $t_j-t_{j-1}$ in the Euclidean metric on $[0,1]$, each such interval is contained in one member of the cover. Then, for each $j\in\{1,\dots,N\}$, choose an open set $U_j\subset M$ with smooth local section \begin{align*} s_j:U_j\to P \end{align*} such that \begin{align*} \gamma([t_{j-1},t_j])\subset U_j. \end{align*} For each $j$, define the local connection form \begin{align*} A_j:TU_j\to\mathfrak{g} \end{align*} by \begin{align*} A_j=(s_j)^*\omega. \end{align*} Thus, for $x\in U_j$ and $v\in T_xU_j$, \begin{align*} A_j(x)(v)=\omega_{s_j(x)}((ds_j)_x(v)). \end{align*} [/step]

[step:Translate horizontality in a trivialization into an ODE on $G$]Fix $j\in\{1,\dots,N\}$. Let $I_j=[t_{j-1},t_j]$. Suppose \begin{align*} \widetilde{\gamma}_j:I_j\to P \end{align*} is a smooth lift of $\gamma|_{I_j}$ whose image lies over $U_j$. Since $s_j$ is a local section, there is a unique smooth curve \begin{align*} g_j:I_j\to G \end{align*} such that \begin{align*} \widetilde{\gamma}_j(t)=s_j(\gamma(t))\cdot g_j(t) \end{align*} for every $t\in I_j$. Let \begin{align*} a_j:I_j\to\mathfrak{g} \end{align*} be the smooth curve defined by \begin{align*} a_j(t)=A_j(\gamma(t))(\dot{\gamma}(t)). \end{align*} Let \begin{align*} \rho:G\times G\to G \end{align*} denote the group multiplication map, $\rho(h,k)=hk$. For each $g\in G$, define the group right translation map \begin{align*} R_g:G\to G \end{align*} by $R_g(h)=hg$. For each $h\in G$, define the principal right action diffeomorphism \begin{align*} \mathcal{R}_h:P\to P \end{align*} by $\mathcal{R}_h(p)=p\cdot h$. Let \begin{align*} \theta^R:TG\to\mathfrak{g} \end{align*} denote the right Maurer-Cartan form, defined by right translation to the identity element: for $g\in G$ and $v\in T_gG$, \begin{align*} \theta^R_g(v)=(dR_{g^{-1}})_g(v). \end{align*} In matrix notation this is $v g^{-1}$. Let \begin{align*} \exp:\mathfrak{g}\to G \end{align*} denote the Lie-group exponential map of $G$. For $h\in G$, let $C_h:G\to G$ denote the conjugation map $C_h(k)=hkh^{-1}$. Define the adjoint representation \begin{align*} \operatorname{Ad}:G\to GL(\mathfrak{g}) \end{align*} by $\operatorname{Ad}_h=(dC_h)_e$, where $e\in G$ is the identity element. For $p\in P$ and $\xi\in\mathfrak{g}$, the fundamental vertical vector generated by $\xi$ at $p$ is the tangent vector at $r=0$ to the curve $r\mapsto p\cdot\exp(r\xi)$. The principal connection form equivariance identity $\mathcal{R}_h^*\omega=\operatorname{Ad}_{h^{-1}}\omega$ for the principal right action and the defining property of $\omega$ on fundamental vertical vectors give \begin{align*} \omega_{\widetilde{\gamma}_j(t)}(\dot{\widetilde{\gamma}}_j(t)) = \operatorname{Ad}_{g_j(t)^{-1}}\bigl(a_j(t)+\theta^R_{g_j(t)}(\dot g_j(t))\bigr). \end{align*} Since $\operatorname{Ad}_{g_j(t)^{-1}}:\mathfrak{g}\to\mathfrak{g}$ is a linear isomorphism, the lift $\widetilde{\gamma}_j$ is horizontal precisely when \begin{align*} \theta^R_{g_j(t)}(\dot g_j(t))=-a_j(t) \end{align*} for every $t\in I_j$. Equivalently, in the tangent bundle of $G$, \begin{align*} \dot g_j(t)=-(dR_{g_j(t)})_e(a_j(t)). \end{align*}[/step]

[guided]We now explain why the horizontal lift problem becomes an ordinary differential equation on the structure group. On the interval $I_j=[t_{j-1},t_j]$, the path $\gamma$ stays inside $U_j$, and $s_j:U_j\to P$ is a smooth local section. Therefore every point of $\pi^{-1}(U_j)$ has a unique expression \begin{align*} p=s_j(x)\cdot g \end{align*} with $x\in U_j$ and $g\in G$. Hence every lift \begin{align*} \widetilde{\gamma}_j:I_j\to P \end{align*} of $\gamma|_{I_j}$ has a unique representation \begin{align*} \widetilde{\gamma}_j(t)=s_j(\gamma(t))\cdot g_j(t) \end{align*} for a smooth curve $g_j:I_j\to G$. The goal is to rewrite the condition \begin{align*} \omega_{\widetilde{\gamma}_j(t)}(\dot{\widetilde{\gamma}}_j(t))=0 \end{align*} as an equation for $g_j$. Define the local connection form \begin{align*} A_j:TU_j\to\mathfrak{g} \end{align*} by $A_j=(s_j)^*\omega$, and define the curve \begin{align*} a_j:I_j\to\mathfrak{g} \end{align*} by \begin{align*} a_j(t)=A_j(\gamma(t))(\dot{\gamma}(t)). \end{align*} This $a_j(t)$ measures the connection form on the velocity of the chosen section lift $s_j(\gamma(t))$. There are two contributions to the derivative of \begin{align*} \widetilde{\gamma}_j(t)=s_j(\gamma(t))\cdot g_j(t). \end{align*} The first contribution comes from moving the base point $s_j(\gamma(t))$ and then applying right translation by $g_j(t)$. The connection form transforms under the principal right action by the principal connection identity \begin{align*} \mathcal{R}_h^*\omega=\operatorname{Ad}_{h^{-1}}\omega. \end{align*} Thus the section contribution gives \begin{align*} \operatorname{Ad}_{g_j(t)^{-1}}a_j(t). \end{align*} The second contribution comes from moving in the group direction. For $h\in G$, let $C_h:G\to G$ be the conjugation map $C_h(k)=hkh^{-1}$, and define the adjoint representation $\operatorname{Ad}:G\to GL(\mathfrak{g})$ by $\operatorname{Ad}_h=(dC_h)_e$, where $e\in G$ is the identity element. For each $g\in G$, let \begin{align*} R_g:G\to G \end{align*} be the group right translation map $R_g(h)=hg$. For each $h\in G$, let \begin{align*} \mathcal{R}_h:P\to P \end{align*} be the principal right action diffeomorphism $\mathcal{R}_h(p)=p\cdot h$. Let \begin{align*} \theta^R:TG\to\mathfrak{g} \end{align*} be the right Maurer-Cartan form, defined by \begin{align*} \theta^R_g(v)=(dR_{g^{-1}})_g(v) \end{align*} for $g\in G$ and $v\in T_gG$. In matrix notation, this is the familiar expression $v g^{-1}$. To derive the vertical contribution, fix $t\in I_j$ and write $x=\gamma(t)$ and $g=g_j(t)$. Let \begin{align*} \exp:\mathfrak{g}\to G \end{align*} denote the Lie-group exponential map of $G$. Define $\xi_t\in\mathfrak{g}$ by \begin{align*} \xi_t=\theta^R_{g_j(t)}(\dot g_j(t)). \end{align*} By the definition of the right Maurer-Cartan form, this means \begin{align*} \dot g_j(t)=(dR_g)_e(\xi_t). \end{align*} Equivalently, the group curve $r\mapsto g_j(t+r)$ has first-order velocity at $r=0$ equal to the velocity of the curve $r\mapsto \exp(r\xi_t)g$. Therefore the group-direction velocity at $s_j(x)\cdot g$ is the tangent vector at $r=0$ to \begin{align*} r\mapsto s_j(x)\cdot \exp(r\xi_t)g. \end{align*} This vector is obtained by first taking the fundamental vertical vector generated by $\xi_t$ at $s_j(x)$ and then applying the principal right translation $\mathcal{R}_g:P\to P$. The defining property of a principal connection form gives value $\xi_t$ on the fundamental vertical vector at $s_j(x)$, and the equivariance identity $\mathcal{R}_g^*\omega=\operatorname{Ad}_{g^{-1}}\omega$ therefore gives the vertical term \begin{align*} \operatorname{Ad}_{g_j(t)^{-1}}\theta^R_{g_j(t)}(\dot g_j(t)). \end{align*} Adding the horizontal-section and vertical-group contributions gives \begin{align*} \omega_{\widetilde{\gamma}_j(t)}(\dot{\widetilde{\gamma}}_j(t)) = \operatorname{Ad}_{g_j(t)^{-1}}\bigl(a_j(t)+\theta^R_{g_j(t)}(\dot g_j(t))\bigr). \end{align*} Because $\operatorname{Ad}_{g_j(t)^{-1}}$ is an isomorphism of $\mathfrak{g}$, this expression vanishes exactly when \begin{align*} a_j(t)+\theta^R_{g_j(t)}(\dot g_j(t))=0. \end{align*} Therefore the lift is horizontal exactly when \begin{align*} \theta^R_{g_j(t)}(\dot g_j(t))=-a_j(t). \end{align*} Since $\theta^R_{g}$ is inverse to $(dR_g)_e:\mathfrak{g}\to T_gG$, this is the ordinary differential equation \begin{align*} \dot g_j(t)=-(dR_{g_j(t)})_e(a_j(t)). \end{align*}[/guided]

[step:Solve the local initial value problem in the structure group] Fix $j\in\{1,\dots,N\}$. For $x\in M$, write $P_x:=\pi^{-1}(\{x\})$ for the fiber of $P$ over $x$. Let $q\in P_{\gamma(t_{j-1})}$ be an initial point. Since $q\in\pi^{-1}(U_j)$, there is a unique element $g_{j,0}\in G$ such that \begin{align*} q=s_j(\gamma(t_{j-1}))\cdot g_{j,0}. \end{align*} Define the smooth time-dependent vector field \begin{align*} V_j:I_j\times G\to TG \end{align*} by \begin{align*} V_j(t,g)=-(dR_g)_e(a_j(t)). \end{align*} For every $(t,g)\in I_j\times G$, the vector $V_j(t,g)$ lies in $T_gG$. Since $a_j:I_j\to\mathfrak{g}$ and the right action differential $(g,\xi)\mapsto(dR_g)_e(\xi)$ are smooth, $V_j$ is smooth. We now justify existence on the whole interval $I_j$, using the special right-invariant form of $V_j$. Let \begin{align*} b_j:I_j\to\mathfrak{g} \end{align*} be the smooth curve $b_j(t)=-a_j(t)$, so that $V_j(t,g)=(dR_g)_e(b_j(t))$. By the local existence and uniqueness theorem for smooth ordinary differential equations, applied in a coordinate chart on the finite-dimensional manifold $G$ to the smooth time-dependent vector field $V_j$, there is a unique smooth maximal solution. We spell out why the maximal solution is defined on the whole compact interval $I_j$. Because $I_j$ is compact and $b_j:I_j\to\mathfrak{g}$ is smooth, the set $b_j(I_j)$ is compact in the finite-dimensional [vector space](/page/Vector%20Space) $\mathfrak{g}$. Choose a coordinate neighbourhood $W_e\subset G$ of the identity element $e\in G$ whose closure is compact and contained in a chart domain, and choose an open neighbourhood $W_e'\subset W_e$ of $e$ whose closure is compact and contained in $W_e$. In these coordinates, the representative of $V_j$ is smooth on the compact set $I_j\times\overline{W_e}$, hence its coordinate expression and first derivatives in the group variable are bounded there. The usual contraction-mapping proof of the local ODE theorem therefore gives a number $\varepsilon_j>0$ such that, for every initial time $\tau\in I_j$ and every initial value in $W_e'$, the solution remains in $W_e$ and exists at least on $I_j\cap(\tau-\varepsilon_j,\tau+\varepsilon_j)$. This is the required uniform local existence time on a fixed compact coordinate neighbourhood. Right translation transfers this same lower bound to every initial point $g_0\in G$. Indeed, the vector field satisfies \begin{align*} V_j(t,hg_0)=(dR_{g_0})_h(V_j(t,h)). \end{align*} Thus the diffeomorphism $R_{g_0}:G\to G$ identifies the equation near $e$ with the equation near $g_0$, so the lower bound $\varepsilon_j$ is independent of the initial group element. If a maximal solution starting at $g_{j,0}$ stopped at a finite right endpoint $\beta<t_j$, choose $\tau<\beta$ with $\beta-\tau<\varepsilon_j$. The uniform local solution starting from the already defined value $g_j(\tau)$ extends the solution beyond $\beta$, contradicting maximality and uniqueness on the overlap. The same argument at a finite left endpoint proves extension to the left. Hence the initial value problem \begin{align*} \dot g_j(t)=V_j(t,g_j(t)),\qquad g_j(t_{j-1})=g_{j,0} \end{align*} has a unique smooth solution on all of $I_j$. Define \begin{align*} \widetilde{\gamma}_j:I_j\to P \end{align*} by \begin{align*} \widetilde{\gamma}_j(t)=s_j(\gamma(t))\cdot g_j(t). \end{align*} Then $\pi(\widetilde{\gamma}_j(t))=\gamma(t)$ because $\pi(s_j(\gamma(t))\cdot g_j(t))=\pi(s_j(\gamma(t)))=\gamma(t)$, and the previous step shows that $\widetilde{\gamma}_j$ is horizontal. Its initial value is $q$ by the choice of $g_{j,0}$. [/step]

[step:Concatenate the local horizontal lifts] We construct $\widetilde{\gamma}$ inductively. On $I_1=[t_0,t_1]$, apply the previous step with $q=p_0$ and obtain a unique horizontal lift \begin{align*} \widetilde{\gamma}_1:I_1\to P \end{align*} with $\widetilde{\gamma}_1(t_0)=p_0$. Assume that $\widetilde{\gamma}_{j-1}$ has been constructed on $I_{j-1}$ for some $j\in\{2,\dots,N\}$. Apply the previous step on $I_j$ with initial point \begin{align*} q=\widetilde{\gamma}_{j-1}(t_{j-1}). \end{align*} This gives a unique horizontal lift \begin{align*} \widetilde{\gamma}_j:I_j\to P \end{align*} with \begin{align*} \widetilde{\gamma}_j(t_{j-1})=\widetilde{\gamma}_{j-1}(t_{j-1}). \end{align*} Define \begin{align*} \widetilde{\gamma}:[0,1]\to P \end{align*} by setting $\widetilde{\gamma}(t)=\widetilde{\gamma}_j(t)$ whenever $t\in[t_{j-1},t_j]$. At common endpoints the adjacent definitions agree by construction. We verify smoothness at each subdivision point. Fix $j\in\{1,\dots,N-1\}$ and set $\tau=t_j$ and $q=\widetilde{\gamma}_j(\tau)=\widetilde{\gamma}_{j+1}(\tau)$. Choose an open neighbourhood $W\subset M$ of $\gamma(\tau)$ and a smooth local section \begin{align*} s:W\to P \end{align*} with $\gamma((\tau-\varepsilon,\tau+\varepsilon)\cap[0,1])\subset W$ for some $\varepsilon>0$; this is possible because $\gamma$ is continuous and $P$ is locally trivial. Let \begin{align*} A:TW\to\mathfrak{g} \end{align*} be the local connection form $A=s^*\omega$, and define \begin{align*} a:(\tau-\varepsilon,\tau+\varepsilon)\cap[0,1]\to\mathfrak{g} \end{align*} by $a(t)=A(\gamma(t))(\dot\gamma(t))$. Let \begin{align*} u_-:(\tau-\varepsilon,\tau]\to G \end{align*} and \begin{align*} u_+:[\tau,\tau+\varepsilon)\to G \end{align*} be the smooth coordinate curves for the two adjacent lifted pieces in this same trivialization. Thus \begin{align*} \widetilde{\gamma}_j(t)=s(\gamma(t))\cdot u_-(t) \end{align*} for $t\in(\tau-\varepsilon,\tau]$, and \begin{align*} \widetilde{\gamma}_{j+1}(t)=s(\gamma(t))\cdot u_+(t) \end{align*} for $t\in[\tau,\tau+\varepsilon)$. The curves $u_-$ and $u_+$ satisfy the same ODE \begin{align*} \dot u(t)=-(dR_{u(t)})_e(a(t)) \end{align*} and have the same value at $\tau$, because both lifted points equal $q$. The local uniqueness theorem for this ODE with initial time $\tau$ gives equality of the two solution germs on a smaller interval around $\tau$. Therefore the two adjacent pieces agree near $\tau$ when expressed in the same smooth trivialization, so the concatenated path is smooth at $t_j$. Thus $\widetilde{\gamma}$ is a smooth horizontal lift of $\gamma$ on $[0,1]$ and satisfies $\widetilde{\gamma}(0)=p_0$. [/step]

[step:Prove uniqueness by reducing any competing lift to the same local ODE] Let $\eta:[0,1]\to P$ be another smooth horizontal lift of $\gamma$ with $\eta(0)=p_0$. We prove $\eta=\widetilde{\gamma}$. On $I_1$, write \begin{align*} \eta(t)=s_1(\gamma(t))\cdot h_1(t) \end{align*} for the unique smooth curve \begin{align*} h_1:I_1\to G. \end{align*} By the ODE computation above, horizontality of $\eta$ implies \begin{align*} \dot h_1(t)=-(dR_{h_1(t)})_e(a_1(t)). \end{align*} The initial condition $\eta(0)=p_0=\widetilde{\gamma}(0)$ gives $h_1(0)=g_1(0)$. By uniqueness for the same initial value problem on $G$, $h_1=g_1$ on $I_1$, so $\eta=\widetilde{\gamma}$ on $I_1$. Inductively, suppose $\eta=\widetilde{\gamma}$ on $[0,t_{j-1}]$. Then $\eta(t_{j-1})=\widetilde{\gamma}(t_{j-1})$. Writing both lifts over $I_j$ in the local section $s_j$ gives two solutions of the same ODE \begin{align*} \dot g(t)=-(dR_{g(t)})_e(a_j(t)) \end{align*} with the same initial value at $t_{j-1}$. ODE uniqueness gives equality on $I_j$. By induction, $\eta=\widetilde{\gamma}$ on every $I_j$, hence on all of $[0,1]$. This proves both existence and uniqueness of the horizontal lift with initial value $p_0$. [/step]