Redundance of the Smoothness of the Inversion Map in the Definition of a Lie Group.

Solution 1:

You are right, the argument isn't conclusive, it's missing a piece, but it's fixable. You need to note that the map $p_1=\pi_1\circ\iota:\Delta'\to G$ has constant rank equal to $\dim G=\dim\Delta'$.

Indeed, let $m=(g,g^{-1})\in\Delta'$ be a point in $\Delta'$, then \begin{array}{cccc} \theta_g&:&G\times G&\longrightarrow & G\times G\\ &&(x,y) & \longmapsto &(gx,yg^{-1}) \end{array} is a diffeomorphism of $G\times G$ and sends $\Delta'$ diffeomorphically to itself (let's call $\widetilde{\theta_g}$ the induced diffeomorphism of $\Delta'$). Then $$p_1\circ\widetilde{\theta_g}=\pi_1\circ\iota\circ\widetilde{\theta_g}=\pi_1\circ\theta_g\circ\iota=L_g\circ\pi_1\circ\iota=L_g\circ p_1$$ (where $L_g$ is left multiplication by $g$). If we calculate the differentials at $m_0=(e,e)\in\Delta'$, we get $$d_{m}p_1\circ d_{m_0}\widetilde{\theta_g}=d_eL_g\circ d_{m_0}p_1 $$ Since $\widetilde{\theta_g}$ and $L_g$ are diffeomorphisms, $p_1$ has the same rank at $m\in\Delta'$ as at $m_0\in\Delta'$, so that it has constant rank ($\leq\dim G$). It follows from the constant rank theorem (see the first theorem in these notes) and the bijectivity of $p_1$, that $p_1$ has constant rank equal to $\dim G$, and now you can apply the global inversion theorem.