Problem: Let $ G $ be a group that is also a smooth manifold, and suppose that the mapping $ (x,y) \mapsto x y $ is smooth. How can we show that $ G $ is a Lie group?
This is a problem from Spivak’s book, and he divides it into three steps:
Find $ f^{-1} $ when $ f: G \times G \to G \times G $ is defined by $ f(x,y) \stackrel{\text{def}}{=} (x,x y) $ for all $ (x,y) \in G \times G $.
Show that $ (e,e) $ is a regular point of $ f $.
Then conclude that $ G $ is a Lie group.
I don’t know how to use (1) to get to (2). How many ways are there to show that a point is regular?
To prove that $(e,e)$ is a regular point, you can easily compute the derivative of $f$ at $(e,e)$ and show that it is invertible.
Next if $(e,e)$ is a regular point, then by the inverse function theorem you know that $f$ is local self diffeomorphism of $G\times G$.
Consequently, $f^{-1}$ is also smooth at $f(e,e)=(e,e)$ and thus $x\mapsto \pi_2\circ f^{-1}(x,e)$ is smooth at $e$ where $\pi_2$ denotes the second projection $(x,y)\mapsto y$.
If you compute this last map, you will get that $x\mapsto x^{-1}$ is smooth at $e$. The final step is to prove or use the following fact:
which comes from the smoothness of the multiplication $(x,y)\mapsto xy$.