[proofplan] We lift the multiplication and inversion of $G$ through the universal covering map after fixing the value over the identity element. The covering-space lifting theorem gives uniqueness because the relevant domains $\widetilde G$, $\widetilde G\times \widetilde G$, and $\widetilde G^3$ are connected and simply connected, while the smooth lifting theorem gives smoothness of the lifted operations. The group axioms are then proved by comparing two lifts of the same map into $G$ which agree at the chosen basepoint. Finally, $p$ is a local diffeomorphism and a Lie [group homomorphism](/page/Group%20Homomorphism), so its differential at the identity is a [Lie algebra](/page/Lie%20Algebra) isomorphism. [/proofplan]

[step:Lift multiplication and inversion through the universal covering map]Let \begin{align*} m:G\times G\to G \end{align*} denote the multiplication map of $G$, and let \begin{align*} \iota:G\to G \end{align*} denote the inversion map of $G$. Since $p:\widetilde G\to G$ is the universal covering map of the connected smooth manifold $G$, the space $\widetilde G$ is connected and simply connected. Hence $\widetilde G\times \widetilde G$ is connected and simply connected. Define the smooth map \begin{align*} F:\widetilde G\times \widetilde G\to G,\qquad F(a,b)=m(p(a),p(b)). \end{align*} Since $\widetilde G\times \widetilde G$ is simply connected and \begin{align*} F(\widetilde e,\widetilde e)=m(e,e)=e=p(\widetilde e), \end{align*} the covering-space lifting theorem gives a unique continuous map \begin{align*} \widetilde m:\widetilde G\times \widetilde G\to \widetilde G \end{align*} such that \begin{align*} p\circ \widetilde m=F \end{align*} and \begin{align*} \widetilde m(\widetilde e,\widetilde e)=\widetilde e. \end{align*} Because $F$ is smooth and $p$ is a smooth covering map, the smooth lifting theorem for covering maps implies that $\widetilde m$ is smooth. Similarly, define \begin{align*} J:\widetilde G\to G,\qquad J(a)=\iota(p(a)). \end{align*} Since $J(\widetilde e)=e=p(\widetilde e)$ and $\widetilde G$ is simply connected, the covering-space lifting theorem gives a unique continuous map \begin{align*} \widetilde\iota:\widetilde G\to \widetilde G \end{align*} such that \begin{align*} p\circ \widetilde\iota=J \end{align*} and \begin{align*} \widetilde\iota(\widetilde e)=\widetilde e. \end{align*} Again, the smooth lifting theorem implies that $\widetilde\iota$ is smooth.[/step]

[guided]The goal is to put a group law on $\widetilde G$ by forcing $p$ to preserve multiplication. Since $p$ is supposed to become a homomorphism, the only possible multiplication on $\widetilde G$ must satisfy \begin{align*} p(\widetilde m(a,b))=p(a)p(b) \end{align*} for all $a,b\in \widetilde G$. Thus we define the map to be lifted by \begin{align*} F:\widetilde G\times \widetilde G\to G,\qquad F(a,b)=m(p(a),p(b)). \end{align*} The covering-space lifting theorem applies because the domain $\widetilde G\times \widetilde G$ is connected and simply connected. The space $\widetilde G$ is connected and simply connected because it is the universal cover of the connected manifold $G$, and the product of simply connected spaces is simply connected. The basepoint condition is also satisfied: \begin{align*} F(\widetilde e,\widetilde e)=m(p(\widetilde e),p(\widetilde e))=m(e,e)=e=p(\widetilde e). \end{align*} Therefore there is a unique lift \begin{align*} \widetilde m:\widetilde G\times \widetilde G\to \widetilde G \end{align*} with \begin{align*} p\circ \widetilde m=F \end{align*} and \begin{align*} \widetilde m(\widetilde e,\widetilde e)=\widetilde e. \end{align*} The smoothness of $\widetilde m$ is not automatic from the topological lifting theorem alone. Here the statement uses the smooth covering-manifold structure on $\widetilde G$, so $p$ is a smooth covering map and hence locally a diffeomorphism. Since $F$ is smooth as a composite of the smooth maps $p\times p$ and $m$, the smooth lifting theorem for covering maps implies that $\widetilde m$ is smooth. The same construction applies to inversion. A homomorphic covering map must satisfy \begin{align*} p(\widetilde\iota(a))=p(a)^{-1}. \end{align*} So we define \begin{align*} J:\widetilde G\to G,\qquad J(a)=\iota(p(a)). \end{align*} The domain $\widetilde G$ is simply connected, and \begin{align*} J(\widetilde e)=\iota(e)=e=p(\widetilde e). \end{align*} Thus the covering-space lifting theorem gives a unique lift \begin{align*} \widetilde\iota:\widetilde G\to \widetilde G \end{align*} with \begin{align*} p\circ \widetilde\iota=J \end{align*} and \begin{align*} \widetilde\iota(\widetilde e)=\widetilde e. \end{align*} Since $J$ is smooth and $p$ is a smooth covering map, the lifted inversion map $\widetilde\iota$ is smooth.[/guided]

[step:Prove associativity by uniqueness of lifts] Define smooth maps \begin{align*} A:\widetilde G^3\to \widetilde G,\qquad A(a,b,c)=\widetilde m(\widetilde m(a,b),c) \end{align*} and \begin{align*} B:\widetilde G^3\to \widetilde G,\qquad B(a,b,c)=\widetilde m(a,\widetilde m(b,c)). \end{align*} The space $\widetilde G^3$ is connected and simply connected. For all $a,b,c\in \widetilde G$, using $p\circ \widetilde m=m\circ(p\times p)$ gives \begin{align*} p(A(a,b,c))=(p(a)p(b))p(c). \end{align*} Associativity in $G$ gives \begin{align*} (p(a)p(b))p(c)=p(a)(p(b)p(c)). \end{align*} Again using $p\circ \widetilde m=m\circ(p\times p)$, \begin{align*} p(a)(p(b)p(c))=p(B(a,b,c)). \end{align*} Hence $A$ and $B$ are lifts of the same map $\widetilde G^3\to G$. Moreover, \begin{align*} A(\widetilde e,\widetilde e,\widetilde e)=\widetilde e=B(\widetilde e,\widetilde e,\widetilde e). \end{align*} By [uniqueness of lifts](/theorems/1885) on the [connected space](/page/Connected%20Space) $\widetilde G^3$, $A=B$. Thus $\widetilde m$ is associative. [/step]

[step:Prove the identity and inverse laws by comparing lifts]Define \begin{align*} L:\widetilde G\to \widetilde G,\qquad L(a)=\widetilde m(\widetilde e,a) \end{align*} and \begin{align*} R:\widetilde G\to \widetilde G,\qquad R(a)=\widetilde m(a,\widetilde e). \end{align*} For every $a\in\widetilde G$, \begin{align*} p(L(a))=p(a) \end{align*} and \begin{align*} p(R(a))=p(a). \end{align*} Thus $L$, $R$, and $\operatorname{id}_{\widetilde G}$ are lifts of the same map $p:\widetilde G\to G$. Since \begin{align*} L(\widetilde e)=\widetilde e=R(\widetilde e)=\operatorname{id}_{\widetilde G}(\widetilde e), \end{align*} uniqueness of lifts on the connected space $\widetilde G$ gives \begin{align*} L=\operatorname{id}_{\widetilde G}=R. \end{align*} Hence $\widetilde e$ is a two-sided identity for $\widetilde m$. Now define \begin{align*} Q:\widetilde G\to \widetilde G,\qquad Q(a)=\widetilde m(\widetilde\iota(a),a) \end{align*} and \begin{align*} S:\widetilde G\to \widetilde G,\qquad S(a)=\widetilde m(a,\widetilde\iota(a)). \end{align*} Let \begin{align*} C:\widetilde G\to \widetilde G,\qquad C(a)=\widetilde e \end{align*} be the constant map. For every $a\in\widetilde G$, \begin{align*} p(Q(a))=p(a)^{-1}p(a)=e=p(C(a)) \end{align*} and \begin{align*} p(S(a))=p(a)p(a)^{-1}=e=p(C(a)). \end{align*} Thus $Q$, $S$, and $C$ are lifts of the same constant map $\widetilde G\to G$ with value $e$. Since \begin{align*} Q(\widetilde e)=\widetilde e=C(\widetilde e)=S(\widetilde e), \end{align*} uniqueness of lifts on the connected space $\widetilde G$ gives \begin{align*} Q=C=S. \end{align*} Therefore $\widetilde\iota(a)$ is a two-sided inverse of $a$ for every $a\in\widetilde G$.[/step]

[guided]The identity law is another lift-uniqueness argument. Consider the left-translation-by-$\widetilde e$ map \begin{align*} L:\widetilde G\to \widetilde G,\qquad L(a)=\widetilde m(\widetilde e,a). \end{align*} Projecting it to $G$ gives \begin{align*} p(L(a))=p(\widetilde m(\widetilde e,a))=p(\widetilde e)p(a)=ep(a)=p(a). \end{align*} So $L$ is a lift of the covering map $p:\widetilde G\to G$. The identity map $\operatorname{id}_{\widetilde G}$ is also a lift of $p$, because \begin{align*} p\circ \operatorname{id}_{\widetilde G}=p. \end{align*} These two lifts agree at the basepoint: \begin{align*} L(\widetilde e)=\widetilde m(\widetilde e,\widetilde e)=\widetilde e=\operatorname{id}_{\widetilde G}(\widetilde e). \end{align*} Since $\widetilde G$ is connected, uniqueness of lifts implies \begin{align*} L=\operatorname{id}_{\widetilde G}. \end{align*} The right identity law is identical with the map \begin{align*} R:\widetilde G\to \widetilde G,\qquad R(a)=\widetilde m(a,\widetilde e). \end{align*} Indeed, \begin{align*} p(R(a))=p(a)p(\widetilde e)=p(a)e=p(a), \end{align*} and $R(\widetilde e)=\widetilde e$, so $R=\operatorname{id}_{\widetilde G}$. For inverses, define \begin{align*} Q:\widetilde G\to \widetilde G,\qquad Q(a)=\widetilde m(\widetilde\iota(a),a). \end{align*} Projection to $G$ gives \begin{align*} p(Q(a))=p(\widetilde\iota(a))p(a)=p(a)^{-1}p(a)=e. \end{align*} Thus $Q$ is a lift of the constant map from $\widetilde G$ to $G$ with value $e$. The constant map \begin{align*} C:\widetilde G\to \widetilde G,\qquad C(a)=\widetilde e \end{align*} is also a lift of that same constant map, since $p(C(a))=p(\widetilde e)=e$. At the basepoint, \begin{align*} Q(\widetilde e)=\widetilde m(\widetilde\iota(\widetilde e),\widetilde e)=\widetilde m(\widetilde e,\widetilde e)=\widetilde e=C(\widetilde e). \end{align*} Uniqueness of lifts on connected domains therefore gives $Q=C$, which is the left inverse identity \begin{align*} \widetilde m(\widetilde\iota(a),a)=\widetilde e. \end{align*} The right inverse identity follows from the same comparison applied to \begin{align*} S:\widetilde G\to \widetilde G,\qquad S(a)=\widetilde m(a,\widetilde\iota(a)). \end{align*} Since \begin{align*} p(S(a))=p(a)p(a)^{-1}=e \end{align*} and \begin{align*} S(\widetilde e)=\widetilde e=C(\widetilde e), \end{align*} we get $S=C$. Therefore every $a\in\widetilde G$ has inverse $\widetilde\iota(a)$.[/guided]

[step:Conclude that the lifted operations define a Lie group and that $p$ is a homomorphism] The previous steps show that $\widetilde m$ is associative, that $\widetilde e$ is a two-sided identity, and that $\widetilde\iota$ gives two-sided inverses. Hence $\widetilde G$ is a group with multiplication $\widetilde m$ and identity $\widetilde e$. Since $\widetilde m$ and $\widetilde\iota$ are smooth, this group structure is a Lie group structure on the smooth manifold $\widetilde G$. For every $a,b\in\widetilde G$, the defining lift identity gives \begin{align*} p(\widetilde m(a,b))=m(p(a),p(b)). \end{align*} Also $p(\widetilde e)=e$. Therefore $p:\widetilde G\to G$ preserves identity and multiplication, so $p$ is a Lie group homomorphism. [/step]

[step:Prove uniqueness of the pointed Lie group structure] Suppose another Lie group structure on the same smooth manifold $\widetilde G$ has identity element $\widetilde e$ and multiplication \begin{align*} \mu:\widetilde G\times \widetilde G\to \widetilde G \end{align*} such that $p$ is a Lie group homomorphism. Then \begin{align*} p(\mu(a,b))=m(p(a),p(b)) \end{align*} for all $a,b\in\widetilde G$, and \begin{align*} \mu(\widetilde e,\widetilde e)=\widetilde e. \end{align*} Thus $\mu$ is a lift of $F=m\circ(p\times p)$ with the same basepoint value as $\widetilde m$. By uniqueness of lifts from the connected space $\widetilde G\times \widetilde G$, one has $\mu=\widetilde m$. If \begin{align*} \nu:\widetilde G\to\widetilde G \end{align*} is the inversion map for that other group structure, then $p\circ\nu=\iota\circ p$ and $\nu(\widetilde e)=\widetilde e$. Hence $\nu$ is the unique lift of $J=\iota\circ p$ with basepoint value $\widetilde e$, so $\nu=\widetilde\iota$. Therefore the Lie group structure is unique. [/step]

[step:Identify the Lie algebra of the universal cover with the Lie algebra of $G$] By construction, $\widetilde G$ is the universal covering space of $G$, so $\widetilde G$ is simply connected. Since $p:\widetilde G\to G$ is a smooth covering map, it is a local diffeomorphism at $\widetilde e$. Therefore its differential \begin{align*} d p_{\widetilde e}:T_{\widetilde e}\widetilde G\to T_eG \end{align*} is a linear isomorphism. The map $p$ is a Lie group homomorphism by the preceding step. By [citetheorem:8803], the [differential of a Lie group homomorphism preserves Lie brackets](/theorems/8803). Thus for all $X,Y\in T_{\widetilde e}\widetilde G$, \begin{align*} d p_{\widetilde e}([X,Y]_{\widetilde{\mathfrak g}})=[d p_{\widetilde e}(X),d p_{\widetilde e}(Y)]_{\mathfrak g}, \end{align*} where $\widetilde{\mathfrak g}=T_{\widetilde e}\widetilde G$ and $\mathfrak g=T_eG$ denote the Lie algebras of $\widetilde G$ and $G$. Since $d p_{\widetilde e}$ is both a linear isomorphism and a bracket-preserving map, it is an isomorphism of Lie algebras. This completes the proof. [/step]