it's true that $\exp(\phi_{*,e}(x)) = \phi(\exp(x))$
My teacher passed on this question but he did not even explain what this "exp" function was, I need this subject to graduate this year, could someone please help? I do not think it is a long or difficult demonstration. Or have a link where you have this demo, if anyone has it, I'll be very grateful
The Lie exponential $\exp:\mathfrak{g}\to G$ is defined by $\exp(x) = \gamma_x (1) $, where $\gamma_x:\Bbb R\to G$ is the $1$-parameter subgroup of $G$ such that $\gamma_x (0) = e$ and $\gamma_x'(0)=x $.
Assume $\phi:G\to H$ is a Lie group homomorphism. By uniqueness, we have that $\phi\circ \gamma_x = \gamma_{\phi_{*,e}(x)} $. Evaluating at $t=1$ gives $$\phi (\exp (x)) = \exp (\phi_{*,e}(x)) $$as wanted.