- #1

Arkuski

- 40

- 0

All there is to show is that the inverse map [itex]i(g)=g^{-1}[/itex] is also a smooth map. We can consider a map [itex]F:G×G\rightarrow G×G[/itex] where [itex]F(g,h)=(g,gh)[/itex] and its inverse is [itex]F^{-1}(x,y)=(x,x^{-1}y)[/itex]. If I show that [itex]F[/itex] is a diffeomorphism, then it should be the case that [itex]i(x)[/itex], which is the second component of [itex]F^{-1}(x,e)[/itex], is smooth.

I have found an answer online that describes [itex]F[/itex] as a very long chain of translations, but can't we simply say that [itex]F=Id_G×L_g[/itex]? And since both [itex]Id_G[/itex] and [itex]L_g[/itex] are diffeomorphisms, isn't their product be a diffeomorphism as well?