Theorem: Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. The exponential map $\exp: \mathfrak{g} \to G$ is a local diffeomorphism in a neighborhood of $0 \in \mathfrak{g}$.

Proof:

We apply the inverse function theorem. To do this, we must show that the differential $d(\exp)_0: \mathfrak{g} \to T_e G$ is an isomorphism.

Step 1: Identify the differential at zero.

For $X \in \mathfrak{g}$, consider the curve $\gamma(t) = \exp(tX)$ in $G$. This is a one-parameter subgroup, and by definition: $\gamma(0) = \exp(0) = e, \quad \gamma'(0) = X$

The differential $d(\exp)_0$ sends $X$ to the tangent vector of $\gamma$ at $t = 0$, which is precisely $X$ when we identify $T_e G \cong \mathfrak{g}$ via left-invariant vector fields.

Step 2: Show $d(\exp)_0 = \text{id}$.

More formally, the exponential map satisfies: $\frac{d}{dt}\bigg|_{t=0} \exp(tX) = X$ for all $X \in \mathfrak{g}$.

Using the definition of the differential, for any $X \in \mathfrak{g}$: $d(\exp)_0(X) = \lim_{t \to 0} \frac{\exp(tX) - e}{t}$

In the identification $T_e G \cong \mathfrak{g}$, this limit is exactly $X$. Therefore: $d(\exp)_0 = \text{id}_{\mathfrak{g}}$

Step 3: Apply the inverse function theorem.

Since $d(\exp)_0: \mathfrak{g} \to T_e G$ is an isomorphism (in fact, the identity under our identification), the inverse function theorem guarantees that there exist neighborhoods $U$ of $0$ in $\mathfrak{g}$ and $V$ of $e$ in $G$ such that: $\exp|_U: U \to V$ is a diffeomorphism.

Step 4: Explicit local inverse.

In the neighborhood where $\exp$ is bijective, the inverse map is called the logarithm: $\log: V \to U$ defined by $\log(\exp(X)) = X$ for $X \in U$.

For matrix groups, this coincides with the matrix logarithm: $\log(g) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(g - I)^n$ which converges for $\|g - I\| < 1$. □

Corollary: Every connected Lie group is generated by any neighborhood of the identity.

Proof: Let $V = \exp(U)$ be a neighborhood of $e$ in $G$ where $U$ is a neighborhood of $0$ in $\mathfrak{g}$. Since $\exp$ is a local diffeomorphism, $V$ contains an open neighborhood of $e$.

Consider the subgroup $H$ generated by $V$. Since $V$ is open, $H$ is open in $G$. But $H$ is also closed (being a subgroup of a Lie group). Since $G$ is connected, the only subset that is both open and closed is $G$ itself. Therefore $H = G$, proving that $V$ generates $G$. □