$Lie(Ker \pi) = Ker d \pi $

Question

I'm trying to understand the proof of

$Lie(Ker \pi) = Ker d \pi $

where

$\pi: G_1 \rightarrow G_2$

is a smooth homomorphism of two Lie groups $G_1, G_2$ and

$d\pi: g_1 \rightarrow g_2$

is it's derivative.

I understand that I can exchange $\pi$ and $d\pi$ in he following sense:

$\pi( exp(tv)) = exp(td\pi(v))$.

What I don't understand is how to prove the following step:

$exp(td\pi(v)) = e \ \forall t \in \mathbb{R} \Rightarrow d\pi(v) = 0$.

Thank you in advance!

Answer 1

You have the sequence of group morphisms $\ker\pi \hookrightarrow G_1 \rightarrow G_2$, where the first morphism is inclusion and the second one is $\pi$. Taking differentials, you obtain $Lie(\ker\pi) \hookrightarrow g_1 \rightarrow g_2$.

The differential of inclusion is injective, and its image is contained in $\ker d\pi$, obviously. Since they have the same dimension, they must be equal. Hence, the differential of inclusion is a natural isomorphism between $Lie(\ker\pi)$ and $\ker d\pi$.