For $GL_n(\mathbb{R})$ and $SL_n(\mathbb{R})$:

The exponential map $\exp: \mathfrak{gl}_n(\mathbb{R}) \to GL_n(\mathbb{R})$ is the standard matrix exponential. For $X \in \mathfrak{gl}_n(\mathbb{R})$: $\exp(X) = \sum_{k=0}^\infty \frac{X^k}{k!}$ This always produces an invertible matrix, with $\det(\exp(X)) = e^{\text{tr}(X)}$.

For $SL_n(\mathbb{R})$, we have $\mathfrak{sl}_n(\mathbb{R}) = \{X : \text{tr}(X) = 0\}$, and $\exp(X) \in SL_n(\mathbb{R})$ precisely when $\text{tr}(X) = 0$.

For $SO(n)$:

The Lie algebra is $\mathfrak{so}(n) = \{X \in M_n(\mathbb{R}) : X^T = -X\}$. For $X \in \mathfrak{so}(n)$: $\exp(X) = I + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \cdots$ satisfies $\exp(X)^T \exp(X) = I$ (orthogonality) and $\det(\exp(X)) = 1$ for connected components.

For $n = 2$, if $X = \begin{pmatrix} 0 & -\theta \\ \theta & 0 \end{pmatrix}$, then: $\exp(X) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ This gives all rotations in the plane, so $\exp: \mathfrak{so}(2) \to SO(2)$ is surjective.

For $SU(n)$:

The Lie algebra is $\mathfrak{su}(n) = \{X \in M_n(\mathbb{C}) : X^* = -X, \text{tr}(X) = 0\}$. For any $X \in \mathfrak{su}(n)$: $\exp(X)^* \exp(X) = I \quad \text{and} \quad \det(\exp(X)) = 1$

For $SU(2)$, writing $X = \begin{pmatrix} ix & y + iz \\ -y + iz & -ix \end{pmatrix}$ with $x, y, z \in \mathbb{R}$: $\exp(X) = \begin{pmatrix} \cos r + i\frac{x}{r}\sin r & \frac{y + iz}{r}\sin r \\ -\frac{y - iz}{r}\sin r & \cos r - i\frac{x}{r}\sin r \end{pmatrix}$ where $r = \sqrt{x^2 + y^2 + z^2}$. The map $\exp: \mathfrak{su}(2) \to SU(2)$ is surjective.

The Heisenberg group: For $X = \begin{pmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} \in \mathfrak{h}$, the exponential is: $\exp(X) = \begin{pmatrix} 1 & a & c + \frac{ab}{2} \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$ The exponential map $\exp: \mathfrak{h} \to H$ is a global diffeomorphism, characteristic of nilpotent groups.