[proofplan] The proof has two directions. First, a discrete central subgroup $\Gamma$ is closed and normal, so the quotient $\widetilde G/\Gamma$ is a Lie group and the quotient map is a covering homomorphism with discrete kernel; this preserves connectedness and the [Lie algebra](/page/Lie%20Algebra). Conversely, if $G$ is any connected Lie group with Lie algebra $\mathfrak g$, the simply connectedness of $\widetilde G$ lets the identity map on $\mathfrak g$ integrate to a homomorphism $\widetilde G\to G$, and this homomorphism is a covering map. Its kernel is discrete and central by [Kernels of Connected Covering Homomorphisms Are Discrete and Central](citetheorem:TEMP-54), so $G$ is the quotient by that kernel. The final uniqueness statement follows by lifting an identity-on-Lie-algebra quotient isomorphism to the identity automorphism of $\widetilde G$. [/proofplan]

[step:Form quotients by discrete central subgroups] Let $\widetilde e\in \widetilde G$ denote the identity element, and let $\Gamma \le Z(\widetilde G)$ be a discrete subgroup. Since $\widetilde G$ is a Lie group, it is Hausdorff, and every discrete subgroup of a Hausdorff Lie group is closed. Since $\Gamma$ lies in the centre $Z(\widetilde G)$, it is normal in $\widetilde G$. Therefore the quotient Lie group \begin{align*} \widetilde G/\Gamma \end{align*} is defined, and the quotient map \begin{align*} q_\Gamma:\widetilde G \to \widetilde G/\Gamma \end{align*} is a smooth surjective [group homomorphism](/page/Group%20Homomorphism). Because $\Gamma$ is discrete, there is an open neighbourhood $U\subset \widetilde G$ of the identity element $\widetilde e$ such that \begin{align*} U\cap \Gamma=\{\widetilde e\}. \end{align*} After replacing $U$ by a smaller symmetric neighbourhood satisfying $U^{-1}U\cap \Gamma=\{\widetilde e\}$, the restriction $q_\Gamma|_U$ is injective. Indeed, if $u_1,u_2\in U$ and $q_\Gamma(u_1)=q_\Gamma(u_2)$, then $u_2^{-1}u_1\in \Gamma$, while $u_2^{-1}u_1\in U^{-1}U$, hence $u_2^{-1}u_1=\widetilde e$ and $u_1=u_2$. By translating this argument, $q_\Gamma$ is locally a diffeomorphism and hence a covering homomorphism. The quotient $\widetilde G/\Gamma$ is connected because it is the continuous image of the [connected space](/page/Connected%20Space) $\widetilde G$. Since $q_\Gamma$ is a local diffeomorphism, its differential at the identity $d(q_\Gamma)_{e_{\widetilde G}}:T_{e_{\widetilde G}}\widetilde G\to T_{q_\Gamma(e_{\widetilde G})}(\widetilde G/\Gamma)$ is a linear isomorphism. Under the given identification $T_{e_{\widetilde G}}\widetilde G=\mathfrak g$, this identifies the Lie algebra of $\widetilde G/\Gamma$ with $\mathfrak g$. [/step]

[step:Integrate the identity Lie algebra map to a covering homomorphism]Let $G$ be a connected Lie group whose Lie algebra is identified with $\mathfrak g$, and let $e_G$ denote its identity element. Define the Lie algebra map \begin{align*} A:\operatorname{Lie}(\widetilde G)&\to \operatorname{Lie}(G) \end{align*} to be the identity map on $\mathfrak g$ under the two prescribed identifications. Since $\widetilde G$ is connected and simply connected, [Lifting Lie Algebra Homomorphisms from Simply Connected Groups](citetheorem:TEMP-44) gives a unique Lie group homomorphism \begin{align*} p:\widetilde G&\to G \end{align*} such that \begin{align*} dp_{\widetilde e}=A. \end{align*} Because $A$ is a linear isomorphism, $dp_{\widetilde e}$ is an isomorphism. For any $\widetilde g\in \widetilde G$, the homomorphism identity \begin{align*} p\circ L_{\widetilde g}=L_{p(\widetilde g)}\circ p \end{align*} shows, after differentiating at $\widetilde e$, that $dp_{\widetilde g}$ is also a linear isomorphism. Hence $p$ is a local diffeomorphism by the [inverse function theorem](/theorems/51). The image $p(\widetilde G)$ is a subgroup of $G$. Since $p$ is a local diffeomorphism, $p(\widetilde G)$ is open in $G$. An open subgroup of a topological group is also closed, because its complement is a union of open cosets. Since $G$ is connected and $p(\widetilde G)$ is nonempty, open, and closed, we get \begin{align*} p(\widetilde G)=G. \end{align*} Thus $p$ is a surjective local diffeomorphism and a Lie group homomorphism. A surjective local diffeomorphism homomorphism is a covering homomorphism.[/step]

[guided]We begin with the identity map on the Lie algebra, but we must turn it into an actual group homomorphism. The relevant result is [Lifting Lie Algebra Homomorphisms from Simply Connected Groups](citetheorem:TEMP-44). Its hypotheses are satisfied because $\widetilde G$ is connected and simply connected, $G$ is a Lie group, and the map $A:\operatorname{Lie}(\widetilde G)\to \operatorname{Lie}(G)$ is a Lie algebra homomorphism: under the given identifications, it is the identity map on $\mathfrak g$, so it preserves the bracket. Therefore there is a unique Lie group homomorphism \begin{align*} p:\widetilde G&\to G \end{align*} with \begin{align*} dp_{\widetilde e}=A. \end{align*} Since $A$ is an isomorphism of finite-dimensional real vector spaces, $dp_{\widetilde e}$ is a linear isomorphism. We now propagate this from the identity to every point of $\widetilde G$. For $\widetilde g\in \widetilde G$, let \begin{align*} L_{\widetilde g}:\widetilde G\to \widetilde G,\quad x\mapsto \widetilde g x \end{align*} and \begin{align*} L_{p(\widetilde g)}:G\to G,\quad y\mapsto p(\widetilde g)y \end{align*} be left translations. The homomorphism property gives \begin{align*} p\circ L_{\widetilde g}=L_{p(\widetilde g)}\circ p. \end{align*} Differentiating this identity at $\widetilde e$ gives \begin{align*} dp_{\widetilde g}\circ d(L_{\widetilde g})_{\widetilde e}=d(L_{p(\widetilde g)})_{e_G}\circ dp_{\widetilde e}. \end{align*} Both left-translation differentials are linear isomorphisms, and $dp_{\widetilde e}$ is a linear isomorphism, so $dp_{\widetilde g}$ is a linear isomorphism. By the inverse function theorem, $p$ is a local diffeomorphism. It remains to see that $p$ reaches all of $G$. Since $p$ is a homomorphism, $p(\widetilde G)$ is a subgroup of $G$. Since $p$ is a local diffeomorphism, $p(\widetilde G)$ contains an open neighbourhood of each of its points, hence is open in $G$. Every open subgroup of a topological group is also closed: the complement is the union of all cosets not equal to the subgroup, and those cosets are open because they are translates of an [open set](/page/Open%20Set). Thus $p(\widetilde G)$ is nonempty, open, and closed in the connected group $G$. Consequently \begin{align*} p(\widetilde G)=G. \end{align*} So $p$ is a surjective local diffeomorphism. For a Lie group homomorphism, this implies the covering property: choose a neighbourhood of the identity on which $p$ is a diffeomorphism onto its image, then translate that neighbourhood by elements of the kernel to obtain the evenly covered neighbourhoods. Hence $p:\widetilde G\to G$ is a covering homomorphism.[/guided]

[step:Identify every connected integration as a quotient] Let \begin{align*} \Gamma:=\ker p=\{\widetilde g\in \widetilde G:p(\widetilde g)=e_G\}. \end{align*} Since $p:\widetilde G\to G$ is a covering homomorphism and $\widetilde G$ is connected, [Kernels of Connected Covering Homomorphisms Are Discrete and Central](citetheorem:TEMP-54) implies that $\Gamma$ is a discrete subgroup of $Z(\widetilde G)$. Define \begin{align*} \Phi:\widetilde G/\Gamma\to G,\quad \widetilde g\Gamma\mapsto p(\widetilde g). \end{align*} This map is well-defined because $p(\widetilde g\gamma)=p(\widetilde g)p(\gamma)=p(\widetilde g)$ for every $\gamma\in\Gamma$. It is a group homomorphism because $p$ is a group homomorphism. It is injective because \begin{align*} \Phi(\widetilde g_1\Gamma)=\Phi(\widetilde g_2\Gamma) \end{align*} implies $p(\widetilde g_2^{-1}\widetilde g_1)=e_G$, hence $\widetilde g_2^{-1}\widetilde g_1\in \Gamma$, so $\widetilde g_1\Gamma=\widetilde g_2\Gamma$. It is surjective because $p$ is surjective. The quotient smooth structure on $\widetilde G/\Gamma$ is the unique smooth structure making the quotient map $q_\Gamma:\widetilde G\to \widetilde G/\Gamma$ a covering homomorphism. Since $p=\Phi\circ q_\Gamma$ and both $p$ and $q_\Gamma$ are local diffeomorphisms, $\Phi$ is a local diffeomorphism. Therefore $\Phi$ is a bijective local diffeomorphism, hence a Lie group isomorphism. Moreover, the identity $p=\Phi\circ q_\Gamma$ and the fact that $d(q_\Gamma)_{e_{\widetilde G}}$ identifies $\operatorname{Lie}(\widetilde G)$ with $\operatorname{Lie}(\widetilde G/\Gamma)$ imply that $d\Phi_{e_{\widetilde G/\Gamma}}$ induces the prescribed identification on $\mathfrak g$. This proves that every connected Lie group with Lie algebra $\mathfrak g$ is isomorphic to $\widetilde G/\Gamma$ for a discrete subgroup $\Gamma\le Z(\widetilde G)$. [/step]

[step:Compare quotients whose isomorphism fixes the Lie algebra] Let $\Gamma_1,\Gamma_2\le Z(\widetilde G)$ be discrete subgroups, and let \begin{align*} q_i:\widetilde G\to \widetilde G/\Gamma_i \end{align*} denote the quotient homomorphism for $i\in\{1,2\}$. First suppose $\Gamma_1=\Gamma_2$. Then the identity map on $\widetilde G$ descends to the identity Lie group isomorphism \begin{align*} \widetilde G/\Gamma_1\to \widetilde G/\Gamma_2 \end{align*} and its differential at the identity is the identity map on $\mathfrak g$ under the natural identifications. Conversely, suppose there is a Lie group isomorphism \begin{align*} F:\widetilde G/\Gamma_1\to \widetilde G/\Gamma_2 \end{align*} whose differential at the identity induces $\operatorname{id}_{\mathfrak g}$. Then \begin{align*} F\circ q_1:\widetilde G\to \widetilde G/\Gamma_2 \end{align*} is a Lie group homomorphism. Its differential at $\widetilde e$ is the same as the differential of $q_2$, under the natural identifications of both quotient Lie algebras with $\mathfrak g$. Since $\widetilde G$ is connected and simply connected, the uniqueness part of [Lifting Lie Algebra Homomorphisms from Simply Connected Groups](citetheorem:TEMP-44) applies to the two homomorphisms $F\circ q_1$ and $q_2$. They have the same differential at the identity, so \begin{align*} F\circ q_1=q_2. \end{align*} Taking kernels gives \begin{align*} \Gamma_1=\ker q_1=\ker(F\circ q_1)=\ker q_2=\Gamma_2, \end{align*} where $\ker(F\circ q_1)=\ker q_1$ because $F$ is an isomorphism. Hence the quotients are isomorphic by an isomorphism inducing the identity on $\mathfrak g$ exactly when the discrete central subgroups are equal. [/step]