The extension of a connected, simply-connected Lie group by $\mathbb{Z}$ is trivial.

Question

I am reading a paper by Calvin Moore on group extensions. On page 54, there is a statement saying that when a Lie group is connected and simply-connected, the the extension of a connected, simply-connected Lie group by $\mathbb{Z}$ is trivial. He said it is easy to see this. However, it is not obvious to me. Is there any reason for that? I am trying not to use the technique of introducing the classifying space since at that time it had not been invented.

Answer 1

Suppose $G$ is a simply connected topological group, $D$ is a discrete group, and we have an extension of topological groups $1\to D \to H \stackrel{p}\to G\to 1$. Then $p$ is a covering map. Explicitly, if $U$ is a neighborhood of $1$ in $H$ which contains no other points of $D$ and $V$ is a neighborhood of $1$ such that $VV^{-1}\subseteq U$, then the cosets of $V$ by elements of $D$ are disjoint (if $aV\cap bV\neq \emptyset$ for $a,b\in D$ then we get that $b^{-1}a$ is equal to an element of $VV^{-1}\subseteq U$ so $a=b$). It follows that $p(V)$ is a neighborhood of $1$ in $G$ which is evenly covered by $p$, and translates of $p(V)$ show that every point of $G$ has an evenly covered neighborhood.

Since $G$ is simply connected, this covering map must be trivial and we have $H\cong G\times D$ (this is an isomorphism of spaces over $G$, but not yet of topological groups). We can then identify $G$ with the connected component of the identity in $H$. Since $D$ is normal in $H$, $G$ acts on it by conjugation, but this action must be trivial since $G$ is connected. So, every element of $G$ commutes with every element of $D$, and we see that $H\cong G\times D$ as a topological group and our extension is trivial.