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.