Let $\mathbb{C} \cup \{\infty\}$ be the Riemann sphere and let$$S^2 = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$$be the unit sphere in $\mathbb{R}^3$. The stereographic projection with center $(0, 0, 1) \in S^2$ is an explicit bijection between the unit sphere and the Riemann sphere provided by the following map:$$p: S^2 \overset{\sim}{\to} \mathbb{C} \cup \{\infty\}, \quad (x, y, z) \mapsto {{x + iy}\over{1 - z}}.$$For any invertible matrix$$g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{GL}_2(\mathbb{C})$$there is an associated fractional linear transformation$$\phi_g: \mathbb{C} \cup \{\infty\} \to \mathbb{C} \cup \{\infty\},$$of the Riemann sphere, given by$$\phi_g(z) := {{az + b}\over{cz + d}} \quad (\text{in particular, we have }\phi_g(\infty) = {a\over c} \text{ and }\phi_g\left(-{d\over c}\right) = \infty, \text{ if }c \neq 0).$$One may use the bijection $p$ to transport the map $\phi_g$ to the unit sphere. That is, we consider the following composite map$$F_g: S^2 \overset{p}{\to} \mathbb{C} \cup \{\infty\} \overset{\phi_g}{\to} \mathbb{C} \cup \{\infty\} \overset{p^{-1}}{\to} S^2.$$Question. If $g \in \text{SU}(\mathbb{C}^2)$, does it follow that the map $F_g$ is a rotation of the sphere $S^2$?
I've attempted this, but I got bogged down with all these maps and get sort of confused. Can anybody help me see this?
Yes! This is the classic "double covering" of $SO(3)$ by $SU(2)$. A good explanation is found at http://people.reed.edu/~jerry/311/rotate.pdf.