Exponential map and the special orthogonal group I need to show that the map exp$: \mathfrak{so}(2) \rightarrow SO(2)$ is surjective.  I already have that $\mathfrak{so}(2) = \{A \in M(2, \mathbb{R}) \ | \ A^T + A = 0\}$ and the map is given by exp$(A) = e^A$.
Any help will be much appreciated.  Thank you
 A: 1) For every anti-symmetric matrix $A$, $(e^A)^T=e^{A^T}=e^{-A}=(e^A)^{-1}$. So $e^A$ is orthogonal and has determinant $\pm 1$. Then $\det e^A=\exp (\rm{trace} A)>0$. Hence the range of your map is contained in the special orthogonal group.
2) Every special orthogonal $2\times 2$ matrix $B$ is of the form 
$$
B=\left( \matrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}\right).
$$
Indeed, the columns have norm $1$ and are orthogonal to each other, by orthogonality of $B$. Starting with the first column, this leaves two possibilities for the second column. The one we have above is the only one which fulfills the positive determinant condition. Now recall that $\cos\theta=\sum_{n\geq 0}\frac{(-1)^n\theta^{2n}}{(2n)!}$ and $\sin\theta=\sum_{n\geq 0}\frac{(-1)^n\theta^{2n+1}}{(2n+1)!}$. Then introduce the matrix
$$
J:=\left( \matrix{0&-1\\ 1&0}\right).
$$
It has order $4$. And straightforward computations show that
$$
B=\sum_{n\geq 0}\frac{\theta^n}{n!}J^n=\sum_{n\geq 0}\frac{(\theta J)^n}{n!}=e^{\theta J}.
$$
So 
$$
A:=\left( \matrix{0&-\theta\\ \theta &0}\right)
$$
is the anti-symmetric log we need to see that $B=e^A$ belongs to the range of your map, which is therefore the whole special orthogonal group.
