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Let $f, g$ be two Lie Group homomorphisms from G to H (smooth group homomorphisms).

Let the induced Lie Algebra homomorphisms $Df(e)$ and $Dg(e)$ be equal. Then if $G$ is connected show that $f=g$.

I know that the set $S=\{x | f(x)=g(x)\}$ is closed. If I show it is open I am done. Now I know $e$ is in $S$. How do I show a nbd is also there?

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  • $\begingroup$ Mm I don't think that there is an easy proof for your statement. You can see Wolfgang Ziller's notes for a proof of this fact easier than for example a classical one that is in Warner's book. $\endgroup$ Commented Feb 8, 2020 at 10:55
  • $\begingroup$ Could you outline the proof briefly? $\endgroup$ Commented Feb 8, 2020 at 10:57

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If $f$ is an homorphism of Lie group, $f(exp(X))=exp(df_{e_G}(X))$, there exists a neighborhood $U$ of $e_G$ such that for every $x\in U$, there exists $X$ in the Lie algebra of $G$ such that $x=exp(X)$, we deduce that $f$ and $g$ coincide in a neighborhood $U$ of $e_G$, since $G$ is connected, $U$ generates $G$ and $f=g$.

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Let $\varphi:G\to H$ a Lie group homomorphism and $d\varphi:\mathfrak{g}\to\mathfrak{h}$ the induced Lie algebra homomorphism, given by $(d\varphi X)_{e_H}=(d\varphi)_{e_G} X_{e_G}$.

Given $\varphi$, let $\Gamma_{\varphi}$ the graph of $\varphi$, namely $\Gamma_{\varphi}=\{(g,\varphi(g))\mid g\in G\}$. You can verify that $\Gamma_{\varphi}$ is an abstract subgroup of $G\times H$ and that $(\Gamma_{\varphi}, \iota)$ (where $\iota$ is the inclusion) is an embedding of $G\times H$ so it is a Lie subgroup of $G\times H$.

What is the Lie algebra of $\Gamma_{\varphi}$ as a subalgebra of the Lie algebra of $G\times H$? Well, you can see that $\mathrm{Lie}(\Gamma_{\varphi})\cong \Gamma_{d\varphi}$, i.e. the graph of the homomorphism $d\varphi$.

Now the proof of your statement:

We have $\mathrm{Lie}(\Gamma_f)\cong \Gamma_{df}=\Gamma_{dg}\cong\mathrm{Lie}(\Gamma_g)$, and if $G$ is connected, then $\Gamma_f$ and $\Gamma_g$ are two connected subgroups which have the same Lie algebra. By uniqueness we must have $\Gamma_f=\Gamma_g$, i.e. $f(a)=g(a)\,\forall a\in G$.

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