In which condition the matrix Lie group $G\simeq \cup_{n=1}^\infty(\exp[g])^n$ is simply connected, where $g$ is a matrix Lie algebra

Question

When is the matrix Lie group $G \simeq \cup_{n=1}^\infty(\exp[\Gamma[g]])^n$ simply connected, where $\Gamma$ is some faithful representation of real or complex Lie algebra $g$.

For example, $su(2)\simeq so(3)$, but the exponential map of definition representation of $su(2)$ is $SU(2)$ which is the universal covering group whose Lie algebra is $su(2)$. The exponential map of adjoint representation of $su(2)$, which is a faithful representation and is $so(3)$, is $SO(3)$ whose fundamental group is $Z_2$.

Another example is $gl(n,R)$ which is isomorphic to $M(n,R)$. The exponential map of definition representation $M(n,R)$ is a local Lie group which can uniquely generate $GL(n,R)$ by $GL(n,R)=\cup_{n=1}^\infty(\exp[M(n,R)])^n$ and $GL(n,R)$ is not simply connected.

So equivalently in which condition the Lie group $G$ generated by $\cup_{n=1}^\infty(\exp[g])^n$ is a universal covering group whose Lie algebra is a matrix Lie algebra $g$, where $\exp[g]$ is image of exponential map of $g$.