# Injective Lie Group Homomorphism is Immersion? [duplicate]

How do I prove the claim of the title?

My definition of Lie Group homomorphism is a smooth group homomorphism. I have to show that $$Df(e)(v)=0$$ will imply $$v=0$$. Viewing this as a derivation, this means $$Df(e)(v)(g)=0$$ for all smooth functions $$g$$ on $$H$$.

Here $$f$$ is the Lie group homomorphism from $$G$$ to $$H$$.

Suppose that $$Df_e(v)=0$$. Then $$(\forall t\in\mathbb R):Df_e(tv)=tDf_e(v)=0$$. And therefore$$(\forall t\in\mathbb R):\exp\bigl(Df_e(tv)\bigr)=e.$$But then$$(\forall t\in\mathbb R):f(e^{tv})=e$$and therefore $$f$$ is not injective.

A different approach to prove this (without the exponential map) is to use the Global Rank Theorem (Lee's book, Smooth manifolds):

Theorem 4.14 (Global Rank Theorem). Let $$M$$ and $$N$$ be smooth manifolds, and suppose $$F:M\to N$$ is a smooth map of constant rank.

(a) if $$F$$ is surjective, then it is a smooth submersion.

(b) if $$F$$ is injective, then it is a smooth immersion.

(c) if $$F$$ is bijective, then it is a diffeomorphism.

Combined with the fact that a Lie group homomorphism $$\varphi:G\to H$$ has constant rank. This fact follows from the equality $$\varphi\circ L_g=L_{\varphi(g)}\circ\varphi$$ where $$g\in G$$ and $$L_g(h)=gh\;\forall\, h\in G$$, and taking the derivative.