# Understanding proof that exponential map of compact connected Lie group is surjective

Let $$G$$ a compact connected Lie group. Then, the exponential map $$\exp: LG \rightarrow G$$ is surjective. (where $$LG$$ is the Lie Algebra of $$G$$).

$$\textbf{Proof:}$$ For any torus $$T' \subset G$$ we have that $$\exp: LT' \rightarrow T'$$ is surjective. So we have that \begin{align} T = \text{im}(\exp: LT \rightarrow T) \subset \text{im}(\exp: LG \rightarrow G) \end{align} Let $$g \in G$$. Then it exists a (maximal) torus $$T \subset G$$ such that $$g \in T$$, so we have that $$g \in \text{im}(\exp)$$ and hence surjectivity. $$\, \, \, _\blacksquare$$

I have some problem to understand how can we say directly that for any torus the exponential map is surjective.

An $$n$$-dimensional torus is a commutative Lie group whose Lie algebra is $$\mathbb{R}^n$$. Consider $$e_1,...,e_n$$ a basis of $$\mathbb{R}^n$$, and $$\Gamma$$ the commutative group generated by $$e_1,...,e_n$$, $$T$$ is the quotient $$\mathbb{R}^n/\Gamma$$. The neutral of $$T$$ is $$p(0)$$, and for every $$u\in\mathbb{R}^n$$, $$exp_0(tu)=p(tu)$$ where $$p$$ is the quotient map $$\mathbb{R}^n\rightarrow T$$. Thus the exponential is just the quotient map $$p$$.