Redundancy of Lie Group Conditions

[Arkuski]

I want to show that if $G$ is a smooth manifold and the multiplication map $m:G×G\rightarrow G$ defined by $m(g,h)=gh$ is smooth, then $G$ is a Lie group.

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

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

 

[quasar987]

But why is L_g smooth in a Lie group if not for the fact that m is smooth?