I am trying to solve the following true or false question:
(True or False) Let $G$ be a connected, compact non-abelian Lie Group $\Longrightarrow$ $\mathrm{exp}:\mathfrak{g} \to G$ is not a local diffeomorphism map.
Since $G$ is compact, then $\mathfrak{g}= \mathfrak{z}(\mathfrak{g})\oplus\mathfrak{s}$ where $\mathfrak{z}(\mathfrak{g})$ is the center of the Lie Algebra $\mathfrak{g}$ and $\mathfrak{s}$ is a semi-simple ideal. Moreover $\mathrm{exp}$ is a surjetive map.
How should I proceed?
It is well known that $$\mathrm{exp}:\mathfrak{g}\to G, $$ satisfies \begin{eqnarray}\mathrm{d}\left(\mathrm{exp}\right)_X = \mathrm{d}L_{e^X}\circ T_X,\quad (*)\end{eqnarray} where $$T_X = \frac{ e^{\mathrm{ad}(X)} -1}{\mathrm{ad}(X)} =\sum_{k\geq0}\frac{1}{(k+1)!}(\mathrm{ad}(X))^k. $$
Since $G$ is compact, all eigenvalues of $\mathrm{ad}(X)$ are purely imaginary numbers, moreover, since $G$ is non-abelian there exists an eigenvalue $i \lambda\neq 0$ of $\mathrm{ad}(X)$ for some $X\in \mathfrak{g}$ and $\lambda \in \mathbb{R}$. Let $c = 2\pi/\lambda $ then $2\pi i$ is an eigenvalue of $\mathrm{ad}(cX)$ $\Rightarrow$ $0$ is eigenvalue of $T_X$ $\Rightarrow$ using $(*)$ $\mathrm{d}(\mathrm{exp})_X$ is not an isomorphism $\Rightarrow$ $\mathrm{exp}$ is not a local diffeomorphism at the point $cX \in \mathfrak{g}$.