Smoothness of the inversion map is redundant in the definition of Lie groups

Question

The question I want to ask is different from this one.

Let $M$ be a smooth manifold which admits a group structure such that the multiplication map $m:G\times G\to G$ defined as $m(g, h)=gh$ for all $g, h\in G$ is a smooth map. Then $G$ is a Lie group.

Hint from professor Lee's book: The map $F\colon G\times G\to G\times G$ defined by $F(g,h) = (g,gh)$ is is a bijective local diffeomorphism.

By taking advantage of the result here(this is the problem 7-2 of Lee's book), I can prove that $dF_{(e,e)}$ is an isomorphism from $T_eG\oplus T_eG$ to itself sending $(X,Y)$ to $(X,X+Y)$, which implies that $F$ is a local diffeomorphism at $(e,e)$. But I don't know how to use this result to obtain that $F$ is local diffeomorphism everywhere. And I also find it hard to prove that $dF_{(g,h)}$ is an isomorphism for general element $(g,h) \in G\times G$.

Thanks in advance!

Answer 1

Let $L_g(h) = gh$ be left-multiplication by $g$, which is a diffeomorphism. Similarly, $R_g(h) = hg$ is a diffeomorphism. You know that $F$ is a diffeomorphism when restricted to some open neighborhood $U$ of $(e,e)$; then $$F|_{(g,h) \cdot U} = (L_g \times L_gR_h) \circ F|_U \circ (L_{g^{-1}} \times R_{h^{-1}}),$$ since the right hand side maps $$(x,y) \mapsto (g^{-1}x, yh^{-1}) \mapsto (g^{-1}x, g^{-1}xyh^{-1}) \mapsto (x, xy),$$ shows that $F$ is a local diffeomorphism on $(g,h) \cdot U$. So it's a bijective map that's a local diffeomorphism everywhere, which means it's a true diffeomorphism.

Answer 2

Here is a straightforward proof. Due to dimensionality, it's enough to show that your map $$F\colon G\times G\rightarrow G\times G,\quad F(g,h)=(g,gh)$$ is an immersion at all points.

Suppose $\mathrm d F_{(g,h)}(v,w)=0$ for some $(v,w)\in T_gG\oplus T_hG$. Since $\mathrm{pr}_1=\mathrm{pr}_1\circ F$, we get $\mathrm d(\mathrm{pr}_1)_{(g,h)}(v,w)=0$ and hence $v=0$. To prove that $w=0$, we consider the equality $m=\mathrm{pr}_2\circ F$, where $m$ denotes the multiplication map; differentiating it yields $\mathrm d m_{(g,h)}(v,w)=0$. On the other hand, $$ \mathrm d m_{(g,h)}(v,w)=\mathrm d(L_g)_h(w)+\mathrm d(R_h)_g(v)=\mathrm d(L_g)_h(w), $$ hence $\mathrm d(L_g)_h(w)=0$, implying $w=0$ since $L_g$ is a diffeomorphism. So $F$ is an immersion at $(g,h)\in G\times G$.

To conclude, since $F$ is a bijection (with inverse $(g,k)\mapsto(g,g^{-1}k)$) and a local diffeomorphism, it is a diffeomorphism, and you can realize the inversion as the composition $\mathrm{inv}=\mathrm{pr_2}\circ F^{-1}\circ \iota$, where $\iota(g)=(g,e)$.