Right, this looks fine to me. I think the more general thing, that if someone would have said to you, you would have understood is the conjugation maps are continuous in topological groups. Indeed, suppose that

is a topological group with multiplication map

and

. Then, the map

given by

is the composition

The first map is just the continuous map

, the second is continuous because each coordinate map is continuous, the third is just the continuous multiplication map, the fourth is just the inclusion

, and the last just multiplication again.

So, if you know that

is a topological group then you're golden because your function

is just

, which is the composition of two continuous functions. But,

is clearly a topological group since the multiplication maps and inversion maps are just rational functions in each coordinate (in fact, this clearly implies that

is a Lie group).