Exponential map of a Lie group

Remarks

Equivalently, $\gamma_X$ is the integral curve through $e$ of the left-invariant vector field with value $X$ at $e$. By rescaling, $\exp(tX) = \gamma_X(t)$ for all $t \in \R$.

Example: for $G = GL_n(\R)$, $\mathfrak{g} = M_n(\R)$ and $\gamma_A(t) = e^{tA}$ is a one-parameter subgroup with $\gamma_A'(0) = A$. So $\exp(A) = e^A$ is the matrix exponential.

Non-example: in $G = SL_2(\R)$, the matrix $\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}$ is not of the form $\exp(X)$ for any $X$ in the Lie algebra $\mathfrak{sl}_2(\R)$. So $\exp$ need not be surjective, even for a connected group.

The map $\exp$ is smooth and its differential at $0$ is the identity of $\mathfrak{g}$. By the inverse function theorem, it restricts to a diffeomorphism from a neighbourhood of $0$ in $\mathfrak{g}$ onto a neighbourhood of $e$ in $G$. For a compact connected Lie group, $\exp$ is surjective.