Question on the connectedness of the orthogonal group

Question

I want to show that the quotient $O_2^- = O_2/SO_2$ is connected. My idea was as follows:

It's easy to show that $SO_2$ is connected. $S0_2$ is a topological group (normal subgroup of a topological group) so we can consider the homeomorphism $\tau : SO_2 \to O_2^-$ defined by $x \mapsto x \sigma$ with $\sigma$ in $SO_2$ (because if we consider the quotient $O_2/SO_2$ we can divise $O_2$ in its equivalence classes $\sigma O_2^-$). We know that a homeomorphism transport the connected property of the group. Thus $O_2^-$ is a connected composant of $O_2$.

Is my reasoning right or are there some mistakes? Thanks.

Answer 1

Apart from the confusion between quotient and complement, the reasoning is mostly correct. It boils down to the follwing :

Let $O_2^-$ be the complement of $SO_2$ in $O_2$. Fix any $\sigma \in O_2^-$. Then $x\in SO_2 \mapsto \sigma x \in O_2^-$ is a homeomorphism. Since $SO_2$ is connected, $O_2^-$ is connected as well.