I know that the composition of covering maps need not be a covering map. But since the usual counter-example is particularly intricate, I wonder if this result is true for Lie groups.
The usual criterion (having finite fibers) surely does not always hold for Lie groups. But perhaps it is still true that the composition of covering maps of Lie groups is still a covering map. I would be glad if anyone has any indication of whether this is true or not.
I haven’t read the note linked by Rob Arthan, but I feel like it’s much too long for the result itself, so I wanted to add a “self-contained” proof. Let $q: Y \rightarrow X, p: Z \rightarrow Y$ be covering maps, write $r=q \circ p: Z \rightarrow X$.
Assume that $X$ is connected, locally path-connected, and semilocally simply connected. Then so are $Y,Z$ and $X$ has a universal cover $W$, write $u: W \rightarrow X$. As $W$ is the universal cover, there is some $q’: W \rightarrow Y$ such that $q \circ q’=u$. It’s not very hard to see that $q’$ is a covering map, and thus that $W$ is the universal cover of $Y$ as well.
By the same reasoning, there is some covering map $p’: W \rightarrow Z$ such that $p \circ p’=q’$. In particular, $r \circ p’=u$, with $p’$ and $u$ covering maps.
The map $r_*: \pi_1(Z) \rightarrow \pi_1(X)$ is a composition of injective maps so is injective. Note that by composition, paths and homotopies lift along $r$. Let $H$ be its image, and let $Z’ \rightarrow X$ be a covering map (witg $s: W \rightarrow Z’$) corresponding to $H$.
Now, it’s easy to see that if $x,x’ \in W$, with $u(x)=u(x’)$, their images in $Z$ are equal iff any path from $x$ to $x’$ is a loop in $Z$, iff the image (under $u$) of some path from $x$ to $x’$ is in the image in $\pi_1(X)$ of the loops of $Z$, iff the image under $u$ of some path from $x$ to $x’$ is in $H \leq \pi_1(X)$. So the images of $x,x’$ in $Z$ are equal iff their images in $Z’$ are equal.
It follows that $W \rightarrow Z \rightarrow X$ and $W \rightarrow Z’ \rightarrow X$ are isomorphic, and thus that $r$ is a covering map.
If $X$ is locally path-connected and semi-locally simply connected and if $p : Z \to Y$ and $q : Y \to X$ are covering projections then $q \circ p : Z \to X$ is also a covering projection. See this paper by Ethan Jerzak for the relevant definitions and the proof. A manifold $X$ satisfies both these conditions (roughly speaking, the latter condition means that every point has a neighbourhood $U$ that contains no loops that cannot be contracted in $X$ - and this is clearly true if $X$ is a manifold). So the composite of any two covering projections $p : Z \to Y$ and $q : Y \to X$ is itself a covering projection, if $X$ is a manifold. Hence, a fortiori, this holds if $X$ is a Lie group.
