Exact non-injectivity of exponential map $su(2) \to SU(2)$ It is well-known that the Lie algebra $su(2)$ has natural basis (as a real vector space) given by the Pauli matrices $\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$, so that a generic element of the corresponding group $SU(2)$ can be written in the form \begin{equation*}
U(\vec{x}) = \exp(i \vec{x} \cdot \vec{\sigma}) \mbox{ for } \vec{x} \in \mathbb{R}^3.
\end{equation*}
This is of course non-injective, since $\exp(i \sigma_z z)$ is invariant under $z \mapsto z + 2 \pi$. Is there a nice characterization of this non-injectivity in general? More specifically, is this a reasonably nice condition on $\vec{x}, \vec{x}'$ that tells whether $U(\vec{x})$ and $U(\vec{x}')$ turn out to be the same group element?
 A: This is a bit tedious because there are cases. I'm going to do this calculation using quaternions. $\mathfrak{su}(2)$ can be thought of as the space of imaginary quaternions $ai + bj + ck$, and the exponential map $\exp : \mathfrak{su}(2) \to SU(2)$ can be computed as exponentiation in the quaternions; it sends imaginary quaternions to unit quaternions. We can write an imaginary quaternion in the form $rq$ where $r \ge 0$ is real and $|q| = 1$ is a unit imaginary quaternion, so $q^2 = -1$, and then we get
$$\exp(rq) = \cos r + q \sin r.$$
So, suppose $r_1 q_1$ and $r_2 q_2$ are imaginary quaternions satisfying $\exp(r_1 q_1) = \exp(r_2 q_2)$. Then the real and imaginary parts must be the same, which, if the imaginary parts are nonzero, gives $\cos r_1 = \cos r_2, |\sin r_1| = |\sin r_2|$, and either $q_1 = q_2$ and $\sin r_1 = \sin r_2$ or $q_1 = -q_2$ and $\sin r_1 = - \sin r_2$. The first case gives $r_1 \equiv r_2 \bmod 2 \pi$ and the second gives $r_1 \equiv -r_2 \bmod 2 \pi$. The signs cancel after multiplying $r_2$ and $q_2$ together, which altogether gives:

Case 1: If $\sin r_1 \neq 0$, then $\exp(r_1 q_1) = \exp(r_2 q_2)$ iff $r_2 q_2 = (r_1 + 2 \pi k) q_1$ for some $k \in \mathbb{Z}$.

This is the generic case, and in particular the set of collisions is discrete. Then there is a special case that happens when $\sin r_1 = \sin r_2 = 0$, which gives $\cos r_1 = \cos r_2$ and places no constraint on $q_1, q_2$. This corresponds to $\exp(r_1 q_1) = \exp(r_2 q_2) = 1$ and we get that either $r_1 \equiv r_2 \equiv 0 \bmod 2 \pi$ (which corresponds to $\exp = 1$) or $r_1 \equiv r_2 \equiv \pi \bmod 2\pi$ (which corresponds to $\exp = -1$). So:

Case 2: If $\sin r_1 = 0$, then $\exp(r_1 q_1) = \exp(r_2 q_2)$ iff $r_1 \equiv r_2 \equiv 0, \pi \bmod 2 \pi$.

In this case the set of collisions is not discrete: there is no condition on $q_1$ and $q_2$ so they can be any imaginary unit quaternions.
A: Qiaochu Yuan kindly provided a careful answer already in terms of quaternion algebra, but by coincidence I simultaneously found the same answer by a different approach: take the usual 2:1 mapping $SU(2) \to SO(3)$ and interpret the latter geometrically. From this we see that $U(\vec{x})$ above corresponds roughly to a rotation about $\hat{n} = \hat{x}$ by an angle $\theta = 2|\hat{x}|$ (in the appropriate double cover sense wherein angles naturally exist modulo $4\pi$, not $2\pi$). Then in most cases, the non-injectivity is generated by the increments \begin{equation*}
\vec{x} \mapsto (1 + 2 \pi / |\vec{x}|) \vec{x}
\end{equation*}
corresponding to $\theta \mapsto \theta + 4 \pi$. On the same geometric grounds, we see that $|\vec{x}| \in \pi \mathbb{Z}$ gives a larger degeneracy since this implies $U(\vec{x}) = \pm 1$ regardless of the orientation of $\hat{x}$; this is exactly Qiaochu Yuan's Case 2.
