Let $G$ be a Lie group and $\phi (\sigma)=\sigma^{-1}$ for each $\sigma\in G$. Let $X$ be a left-invariant vector field, i.e. $$dl_\sigma\circ X = X\circ l_\sigma$$ I wonder why $d\phi(X)$ is right-invariant, that is $$\text{why }\ dr_\sigma \circ d\phi(X) = d\phi(X)\circ r_\sigma\ ?$$
I'm asking this because I am trying to solve exercise 16 of chapter 3 of the book 'Foundations of Differentiable Manifolds and Lie Groups', by F. Warner.
Thanks in advance. Any help is welcome!
To say $d\phi(X)$ is right-invariant is to say that for any $\sigma\in G,dr_{\sigma}(d\phi(X))=d\phi(X).$ Equivalently, since $G$ is a group, it suffices to prove that $dr_{\sigma^{-1}}(d\phi(X))=d\phi(X)$ and this is easy once we notice that $\phi\circ l_{\sigma}(g)=g^{-1}\sigma^{-1}=\phi(g)\phi(\sigma)=r_{\phi(\sigma)}\phi(g)=(r_{\sigma^{-1}}\circ \phi)(g)$. This fact, and the left-invariance of $X$ imply that $d\phi(X)=d(\phi\circ l_{\sigma})(X)=d(r_{\sigma^{-1}}\circ \phi)(X)=dr_{\sigma^{-1}}(d\phi(X))$