If $\varphi: G \rightarrow G$ is a diffeomorphism defined by $\varphi(\sigma)=\sigma^{-1}$ and $X$ is a left-invariant vector field on $G$, then $d\varphi(X)$ is the right-invariant vector field whose value at $e$ is $-X(e)$.
This question is from F. Warner's Manifolds, Chapter 3, Problem 16, and was answered here. I am trying to understand the solution presented in that answer, however.
According to the definition in the book, we are to show that $$dr_\sigma \circ d\phi(X) = d\phi(X)\circ r_\sigma.$$ On the other hand, it suffices to show that $$dr_{\sigma}(d\phi(X))=d\phi(X)$$ according to that answer. Why can we remove the $r_\sigma$ from the RHS? Moreover, why is it true that $$d\phi(X)=d(\phi\circ l_{\sigma})(X)$$ as claimed in the answer?
Many thanks — I have already spent too long on this problem!