Lie Groups Homomrphism Question (Tangent Space)

Question

I need to solve this:

If $f:G\rightarrow H $ is a Lie group homomorphism, then $T_{e_{1}}\ker(f)=\ker(df_{e_{1}})$

Here $e_{1}$ denotes the identity of $G$ and $df_{e_{1}}$ its the differential of $f$ at $e_{1}$.

Remembering that a Lie group homomorphism is a differentiable map ($C^{\infty}$) that is also a group homomorphism.

If anyone can help me, I'll be thankful.

Answer 1

Recall that you have the following commutative diagram :

$\require{AMScd} \begin{CD} T_eG @>{df_e}>> T_eH\\ @V{\exp}VV @VV{\exp}V\\ G @>>{f}> H \end{CD}$

Now if $df_e(X)=0$, then $df_e(tX) = 0$ for all $t$ so that $\exp(df_e(tX)) = 1$, therefore by commutativity of the above, $f(\exp(tX))=1$, so $\exp(tX)\in \ker(f)$ for all $t$; therefore $X\in T_e\ker(f)$.

Conversely, if $X\in T_e\ker(f)$, you can differentiate the equation $f(\exp(tX))=1$ at $t=0$ to get $df_e(X)=0$, so $X\in \ker(df_e)$.