Suppose $H$ is a closed subgroup of a Lie group $G$. Then in Lee's book [Introduction to Smooth Manifolds](https://doi.org/10.1007/978-0-387-21752-9) (Ch. 9) he showed that the action $H\times G\to G$ $(h,g)\mapsto gh$ is a smooth, free, proper action. I have a small problem regarding showing the action is proper. Note that if $g\_ih\_i\to g\_0$ and $g\_i\to g\_{00}$ then we must have that $h\_i\to g\_{00}^{-1}g\_0\mathrel{:=}h\_0$ and as $H$ is closed $h\_i\to h\_0\in H$. I want to understand why this would imply the action is proper. I actually need to show that $h\_i\to h$ in the topology of $H$ but the above argument shows that the convergence happens in the subspace topology of $H$. I know that as $H$ is closed, $H$ must be embedded. But to this point in Lee's book this is not proved. Is there any way to see why the argument in the Lee's book makes sense?



A question regarding the action of a Lie subgroup
As @Ramiro Lafuente pointed out in a [comment](https://mathoverflow.net/questions/378266/a-question-regarding-the-action-of-a-lie-subgroup#comment960592_378266), there's a gap in this proof.


You're apparently using the first edition of my *Smooth Manifolds* book. The problem is fixed in the second edition, because the closed subgroup theorem is proved before this theorem.


Meanwhile, some years ago I posted a correction in my errata list for the first edition, which you can find [here](https://sites.math.washington.edu/%7Elee/Books/ISM/).


