kernel of the exponential map is isomorphic to the singular homology group

Let $G$ be an algebraic torus or an abelian variety over the complex numbers. Then $G(\mathbb{C})$ is a complex Lie group. Is it true that we have the following exact sequence ?

$0 \to H_1(G(\mathbb{C}),\mathbb{Z}) \xrightarrow{\alpha} Lie(G(\mathbb{C})) \xrightarrow{\exp} G(\mathbb{C}) \to 0$

Examples :

1) If $G=\mathbb{C}^{*}$, the exact sequence is just

$0 \to \mathbb{Z} \xrightarrow{2\pi i} \mathbb{C} \xrightarrow{\exp} \mathbb{C}^{*} \to 0$

2) If $G$ is an abelian variety of dimension $d$, then $G(\mathbb{C})=\mathbb{C}^n/\Lambda$ where $\Lambda$ is a maximal lattice of $\mathbb{C}^n$. The exact sequence is

$0 \to \Lambda \to \mathbb{C}^n \to \mathbb{C}^n/\Lambda \to 0$

Could you please also explain to me why the map $\alpha$ and the exponential map are what they are in the above cases ? Moreover what are their descriptions in general ? I think I know the description of the exponential map in general, but I don't understand why it reduces to the quotient map in example 2. Thank you very much.

Any real or complex Lie group $G$ has a universal cover $\widetilde{G}$ on which the fundamental group $\pi_1(G)$ acts with quotient $G$. This always comes from a short exact sequence
$$1 \to \pi_1(G) \to \widetilde{G} \to G \to 1.$$
Because $G$ is a topological group, the natural map $\pi_1(G) \to H_1(G)$ is an isomorphism, and if $G$ is abelian, then the exponential map $\mathfrak{g} \to \widetilde{G}$ is also an isomorphism.
• Thank you for your answer. I wonder if you could give me reference for the statement " If $G$ is abelian, $\mathfrak{g}\to \tilde{G}$ is an isomorphism." – raynor14 Feb 18 '15 at 0:23
• @raynor: if $G$ is abelian, then the exponential map is a homomorphism, and $\mathfrak{g}$ is simply connected (above my Lie groups are always connected), so it must be the universal cover. – Qiaochu Yuan Feb 18 '15 at 8:36