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In my notes I have the following claim.

Let $G$ be a simply connected abelian Lie group. Since $G$ is abelian, we have that $exp(A+B) = exp(A) exp(B)$ for all $A,B∈Lie(G)$. Therefore, the exponential map is an isomorphism between the additive group $Lie(G)$ and $G$.

I think I understand surjectivity because we know $exp : Lie(G) \to G$ restricts to a diffeomorphism on a neighbourhood $U\ni 0$. And since $G$ is connected, we have that $\langle U \rangle=G$ so any element $g\in G$ can be written $g=exp(X_1)^{\pm 1}\ldots exp(X_n)^{\pm 1}=exp(\pm X_1 \pm \ldots \pm X_n)$ (because $G$ is $G$ is abelian so $Lie(G)$ is abelian as well). But why is it injective and what does it has to do with simple connectivity?

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    $\begingroup$ I'd try to argue that $\exp: \mathfrak{g}\to G$ is the universal cover of $G$ (for example, see this). But $G$ is assumed simply connected. So... $\endgroup$
    – Ivo Terek
    Commented Jul 17, 2020 at 23:29

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The lie algebra $\mathfrak{g}$ is an abelian Lie group, and as you pointed out, $\exp:\mathfrak{g}\to G$ is a homomorphism. Since there is a neighbourhood of $0$ on which $\exp$ is a diffeomorphism, the kernel $\ker (\exp)$ of this homomorphism is discrete. Therefore $\exp$ is a covering map. Since $G$ is simply-connected, it must be a homeomorphism, and in particular bijective. So it is an isomorphism.

EDIT. Regarding the discreteness of the kernel:

Let $U\ni 0 $ be a neighbourhood of zero such that $\exp:U\to \exp(U)$ is a diffeomorphism. In particular, for $v\in U\setminus\{0\}%$, we have $\exp(v)\neq\exp(0)=1$, since $\exp$ is injective on $U$. So $K\cap U = \{0\}$. But $K\cap U$ is open in $K$ in the subspace topology by definition, so $\{0\}$ is open in $K$, which means $K$ is discrete.

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    $\begingroup$ Thank you. Why the fact that there is a neighbourhood of $0$ on which $\operatorname{exp}$ is a diffeomorphism implies that the kernel is discrete? $\endgroup$
    – roi_saumon
    Commented Jul 18, 2020 at 16:18
  • $\begingroup$ @roi_saumon Let $U\ni 0 $ be a neighbourhood of zero such that $\exp:U\to \exp(U)$ is a diffeomorphism. In particular, for $v\in U\setminus\{0\}%$, we have $\exp(v)\neq\exp(0)=1$, since $\exp$ is injective on $U$. So $K\cap U = \{0\}$. But $K\cap U$ is open in $K$ in the subspace topology by definition, so $\{0\}$ is open in $K$, which means $K$ is discrete. $\endgroup$
    – Cronus
    Commented Jul 18, 2020 at 16:53
  • $\begingroup$ Oh, nice! I didn't knew that you just need to show that ${0}$ is open because topological group are homogeneous spaces. $\endgroup$
    – roi_saumon
    Commented Jul 18, 2020 at 17:05
  • $\begingroup$ @Roi_Saumon right, sorry, perhaps I ought to have mentioned that. $\endgroup$
    – Cronus
    Commented Jul 18, 2020 at 17:07

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