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

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 $(g,h) \in G\times G$.

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.
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 $$\iota(g)=(g,e)$$.