Proposition 6

Let G be a connected and coverable topological group. There exist a topological group $\widetilde{G}$, with identity element $\widetilde{e}$, and a continuous homomorphism $p:\widetilde{G}\rightarrow G$ satisfying the following assertions:

a) The space $\widetilde{G}$ is simply connected and simply connected by arcs. Endowed with the mapping $p$, the pointed space $(\widetilde{G},\widetilde{e})$ is a universal covering of the pointed space $(G, e)$.

(b) The kernel N of $p$ is a discrete subgroup of $\widetilde{G}$, contained in the center of $\widetilde{G}$. The group homomorphism $N\rightarrow$ Aut$_G(\widetilde{G})$ which, to $n\in N$, associates the right translation in $\widetilde{G}$, is a group isomorphism. The homomorphism of $\pi_1(G)$ into N which, to $\gamma \in \pi_1(G)$, associates the unique element $n$ of N such that $\widetilde{e}\cdot \gamma =n$, is a group isomorphism.

(c) If $G'$ is a topological group, with identity element $e'$, and if $p': G'\rightarrow$ G is a group homomorphism which makes $G'$ into a covering of G, the unique continuous mapping $u:\widetilde{G}\rightarrow G'$ such that $u(\widetilde{e}) =e'$ and $p'\circ u=p$ is a group homomorphism. Endowed with the mapping $u,(\widetilde{G},\widetilde{e})$ is a universal covering of $(G', e')$.

By IV, p. 342, Theorem 1, there exists a covering $(\widetilde{G}, p)$ of G, simply connected and simply connected by arcs, Galoisian with group $\pi_1(G)^{\circ}$, and a point $\widetilde{e}$ of $\overset{-1}{p}(e)$ such that the pointed covering $(\widetilde{G},\widetilde{e})$ is a universal covering of $(G, e)$.

By IV, p. 375, Prop. 4 and IV, p. 377, Prop. 5, there exists on the space $\widetilde{G}$ a unique group structure compatible with its topology for which $p$ is a group homomorphism and $\widetilde{e}$ an identity element. It follows from Prop. 4 that the group $\widetilde{G}$ satisfies assertions (a) and (b).

Let us prove assertion (c). Let $G'$ be a topological group and let $p': G'\rightarrow G$ be a group homomorphism which makes $G'$ into a covering of G. Let $e'$ be the identity element of $G'$. The existence and uniqueness of a mapping $u:\widetilde{G}\rightarrow G'$ such that $p=p'\circ u$ and $u(\widetilde{e}) =e'$ follow from the fact that $(\widetilde{G},\widetilde{e})$ is a universal covering of the pointed space $(G, e)$. Since the mappings $p$ and $p'$ are surjective group homomorphisms, $u$ is a group homomorphism. Endowed with the mapping $u,\widetilde{G}$ is a covering of $G'($I, p. 81, Prop. 7), hence $(\widetilde{G},\widetilde{e})$ is a universal covering of $(G', e')$ (IV, p. 345, Cor. 2 of Theorem 1).

Under the hypotheses of the proposition, we shall say, by abuse, that $\widetilde{G}$ is a universal covering of G.