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

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:

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

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 $$\ker d\pi$$.
• Thank you for the response. I'm not sure I fully understand though. Can we assume the differential of the inclusion is injective since the inclusion is the identity into it's image? Also I don't see where we get that the image of $Lie(ker\pi)$ is in $kerd\pi$. Commented Aug 17, 2023 at 9:23
• @diesmond About the first question: the differential of any injective smooth map is also injective. About the second one: you know that $\pi\circ i = 1$, the constant $1$ map. -The differential of a constant map is $0$, so $0 = d(\pi\circ i) = d\pi \circ di$, which tells us clearly that any vector in the image of $di$ is going to be in $\ker d\pi$ (the composition of both maps is 0) Commented Aug 17, 2023 at 9:52