A Lie group $ G $ is an $r$-times-differentiable manifold endowed with a group structure, i.e. with an associative binary operation
$$ \mu:\quad G\times G \longrightarrow G :\qquad\left\{x\,, y\right\} \longmapsto x\centerdot y $$
and an inversion operation
$$ \zeta:\quad G \longrightarrow G :\qquad g \longmapsto g^{-1}~~, $$
both of which are $r$-times-differentiable.
Now take the inverses of all points residing in an open set $U$: $$ \tilde{U}\equiv\zeta(U)~~. $$ How to show that $\tilde{U}$ is open, i.e. that the inversion $\zeta$ is continuous?
To be more specific, consider a homeomorphism $$ \alpha:\quad U\subseteq G\;\longrightarrow\;W\subseteq{\mathbb{R}}^N~~. $$
If $\tilde{U}$ too is open (which I wish to prove), there should exist a homeomorphism $$ \beta:\quad \tilde{U}\subseteq G\;\longrightarrow\;W\subseteq{\mathbb{R}}^N~~. $$
By saying that the inversion $\zeta$ is differentiable, we actually imply that differentiable is $$ \beta^{-1}\circ\zeta\circ\alpha~~. $$ Is that correct?
Would it be right to say that, by assuming the differentiability, I thereby imply the existence of $\beta$ and, thence, the fact that $\tilde {U}$ is open?
A differentiable function is continuous (standard analysis fact) and so $\zeta$ (being self-inverse) is a homeomorphism and so an open map.