If $G$ is a matrix group and $\U0001d524$ is its Lie algebra then there are neighbourhoods ${U}^{\prime}\subset \U0001d524$ of the zero matrix and ${V}^{\prime}\subset G$ of the identity such that $\mathrm{exp}({U}^{\prime})={V}^{\prime}$ and ${\mathrm{exp}}_{{U}^{\prime}}:{U}^{\prime}\to {V}^{\prime}$ is invertible.
Optional: Local exponential charts
Local coordinates for GL(n,R)
The exponential map $\mathrm{exp}:\U0001d524\U0001d529(n,\mathbf{R})\to GL(n,\mathbf{R})$ is not invertible, but we have seen that there are neighbourhoods $U\subset \U0001d524\U0001d529(n,\mathbf{R})$ of the zeromatrix and $V\subset GL(n,\mathbf{R})$ of the identity matrix such that $\mathrm{exp}(U)=V$ and ${\mathrm{exp}}_{U}:U\to V$ is invertible, with inverse $\mathrm{log}:V\to U$ .
We can think of this as providing for us coordinates near $I\in GL(n,\mathbf{R})$ , namely $\mathrm{exp}\left(\begin{array}{ccc}\hfill {a}_{11}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {a}_{1n}\hfill \\ \hfill \mathrm{\vdots}\hfill & \hfill \hfill & \hfill \mathrm{\vdots}\hfill \\ \hfill {a}_{n1}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {a}_{nn}\hfill \end{array}\right)$ gives a parametrisation of $V$ , so we can think of the matrix entries ${a}_{ij}$ as coordinates on $V$ : anything in $V$ can be written in this form for a unique collection of numbers ${a}_{ij}$ .
Local coordinates for matrix groups
We could like the same to work for any matrix group $G\subset GL(n,\mathbf{R})$ . Namely, we would like to show:
Let ${U}^{\prime}=U\cap \U0001d524$ and ${V}^{\prime}=V\cap G$ . First, we note that $\mathrm{exp}$ does indeed go from $\U0001d524$ to $G$ by definition of $\U0001d524$ , and $\mathrm{exp}(U)=V$ , so $\mathrm{exp}({U}^{\prime})\subset {V}^{\prime}$ .
So what is left to prove? The map $\mathrm{exp}:U\to V$ is invertible, hence injective, so its restriction ${\mathrm{exp}}_{{U}^{\prime}}$ must also be injective. But is ${\mathrm{exp}}_{{U}^{\prime}}$ surjective? In other words, is it clear that $\mathrm{log}({V}^{\prime})\subset {U}^{\prime}$ ?
This figure shows what you might imagine going wrong (we will later show that this doesn't happen, at least if you shrink $U$ and $V$ ). It shows a cartoon of a subgroup $G\subset GL(n,\mathbf{R})$ which "wraps back on itself" and gets very close to the identity but never quite gets there. You can now imagine that when you intersect with a very small $V$ (in red), $\mathrm{exp}({U}^{\prime})$ (in blue) could end up missing this appendage of $G$ which wraps back towards the identity (in green), because to get to this appendage you have to exponentiate something very large. In other words, $\mathrm{exp}({U}^{\prime})$ is the blue bit and ${V}^{\prime}$ is everything which is blue or green. In the end, we will show this doesn't happen, so this is a cartoon picture of something which doesn't happen. You therefore shouldn't be too annoyed if the picture doesn't make sense.
If surjectivity of ${\mathrm{exp}}_{{U}^{\prime}}$ fails then there's an element $g\in {V}^{\prime}=G\cap V$ such that $g\notin \mathrm{exp}({U}^{\prime})$ . We could try to fix this by shrinking $U$ and $V$ , but let's suppose that doesn't help us. This will mean there is a sequence ${g}_{i}\in {V}^{\prime}$ such that ${g}_{i}\to I$ and ${g}_{i}\notin \mathrm{exp}({U}^{\prime})$ for all $i$ (you should imagine a sequence of matrices on the green appendage, tending to the origin in the picture).
Let's assume that there is such a sequence and aim to derive a contradiction.

Recall that $\U0001d524\subset \U0001d524\U0001d529(n,\mathbf{R})$ is a subspace. Pick a vector space complement $W$ for $\U0001d524$ , that is $\U0001d524\cap W=\{0\}$ and $\U0001d524+W=\U0001d524\U0001d529(n,\mathbf{R})$ . In the figure, $\U0001d524$ is supposed to be the horizontal axis (tangent to $G$ at the identity) and $W$ is supposed to be the vertical axis.
I claim that the map $F:\U0001d524\oplus W\to GL(n,\mathbf{R})$ defined by $F(v,w)=\mathrm{exp}(v)\mathrm{exp}(w)$ is locally invertible like $\mathrm{exp}$ , i.e. there exists a neighbourhood ${N}_{1}$ of $0\in \U0001d524\oplus W$ and a neighbourhood ${N}_{2}$ of $I\in GL(n,\mathbf{R})$ such that $F({N}_{1})={N}_{2}$ and $F:{N}_{1}\to {N}_{2}$ is invertible. This is proved using the inverse function theorem, just like for $\mathrm{exp}$ (compute the derivative of $F$ at the zero map and show this derivative is invertible). I leave it as an exercise to fill in the details.
Therefore, if ${g}_{i}$ is sufficiently close to the identity (which it is for large $i$ ), then ${g}_{i}=\mathrm{exp}({v}_{i})\mathrm{exp}({w}_{i})$ for some sequence ${v}_{i}\in \U0001d524$ and ${w}_{i}\in W$ .

Our sequence ${g}_{i}$ is not in $\mathrm{exp}({U}^{\prime})$ , so ${w}_{i}\ne 0$ for all $i$ . In particular, we can divide ${w}_{i}$ by its matrix norm to get a matrix ${w}_{i}/\parallel {w}_{i}\parallel \in W$ with norm 1. Since the set of matrices with norm 1 is closed and bounded (compact), the sequence ${w}_{i}/\parallel {w}_{i}\parallel \in W$ converges to some matrix $w\in W$ with norm 1 (in particular, $w\ne 0$ ).

We are going to prove that $w\in \U0001d524$ ; this will give us a contradiction, as $w\in W$ and $W$ is a complement for $\U0001d524$ . For this, we need to show that $\mathrm{exp}(tw)\in G$ for all $t\in \mathbf{R}$ .
Fix $t$ . Consider $t/\parallel {w}_{i}\parallel $ and take its integer and fractional parts $t/\parallel {w}_{i}\parallel ={n}_{i}+{\u03f5}_{i}$ , ${n}_{i}\in \mathbb{N}$ , $\u03f5\in [0,1)$ . Since ${g}_{i}\to I$ as $i\to \mathrm{\infty}$ , we know that ${w}_{i}\to 0$ , so $\parallel {w}_{i}\parallel \to 0$ , hence $t/\parallel {w}_{i}\parallel \to \mathrm{\infty}$ (as $t$ is fixed) and hence ${n}_{i}\to \mathrm{\infty}$ .

We want to show that $\mathrm{exp}(tw)\in G$ . First note that $\mathrm{exp}({w}_{i})\in G$ for all $i$ because $\mathrm{exp}({w}_{i})=\mathrm{exp}({v}_{i}){g}_{i}$ and $\mathrm{exp}({v}_{i})$ and ${g}_{i}$ are both in $G$ . Because $G$ is a group, ${(\mathrm{exp}({w}_{i}))}^{{n}_{i}}\in G$ and ${(\mathrm{exp}({w}_{i}))}^{{n}_{i}}=\mathrm{exp}({n}_{i}{w}_{i})$ because ${w}_{i}$ commutes with itself. The sequence $\mathrm{exp}({n}_{i}{w}_{i})=\mathrm{exp}(t{w}_{i}/\parallel {w}_{i}\parallel {\u03f5}_{i}{w}_{i})$ converges to $\mathrm{exp}(tw)$ because ${w}_{i}/\parallel {w}_{i}\parallel \to w$ and ${\u03f5}_{i}{w}_{i}\to 0$ .
Because $G$ is topologically closed, this limit lives in $G$ , so $\mathrm{exp}(tw)\in G$ . This argument works for every $t$ , so we're done.
The outcome of all this is that, via the exponential map, local coordinates on $\U0001d524$
near the zero matrix give us local coordinates on $G$
near the identity. We call this an