The strategy is to show that for any ordered triple $(z_0, z_1, z_\infty)$ of distinct points, there is a unique Möbius map sending $z_0 \mapsto 0$, $z_1 \mapsto 1$, $z_\infty \mapsto \infty$. Then the general case follows by composing two such maps. **Step 1: Construct $g$ sending $(z_0, z_1, z_\infty) \mapsto (0, 1, \infty)$.** If $z_0, z_1, z_\infty \neq \infty$, define: \begin{align*} g(z) = \frac{(z - z_0)(z_1 - z_\infty)}{(z - z_\infty)(z_1 - z_0)}. \end{align*} One verifies $g(z_0) = 0$, $g(z_1) = 1$, $g(z_\infty) = \infty$, and the determinant condition holds since the three points are distinct. The cases $z_0 = \infty$, $z_1 = \infty$, or $z_\infty = \infty$ are handled by appropriate simplifications (omitting the corresponding factors). **Step 2: Existence for the general case.** Similarly construct $h$ sending $(w_0, w_1, w_\infty) \mapsto (0, 1, \infty)$. Then $f = h^{-1} \circ g$ satisfies $f(z_i) = w_i$ for $i = 0, 1, \infty$. **Step 3: Uniqueness.** Suppose $f, f' \in \mathcal{M}$ both satisfy $f(z_i) = f'(z_i) = w_i$. Then $\varphi = g \circ f^{-1} \circ f' \circ g^{-1}$ fixes $0$, $1$, and $\infty$. Writing $\varphi(z) = \frac{az + b}{cz + d}$: the condition $\varphi(\infty) = \infty$ gives $c = 0$; $\varphi(0) = 0$ gives $b = 0$; $\varphi(1) = 1$ gives $a = d$. So $\varphi = \mathrm{id}$, hence $f = f'$.