The strategy is to show existence and uniqueness of the Möbius map separately. For existence, we construct auxiliary maps sending each triple to $\{0, 1, \infty\}$ and compose. For uniqueness, we show any Möbius map fixing $0, 1, \infty$ must be the identity. **Step 1: Existence — construct the map.** Given distinct $z_0, z_1, z_\infty \in \mathbb{C}_\infty$, define $g \in \mathcal{M}$ by: \begin{align*} g(z) = \frac{(z - z_0)(z_1 - z_\infty)}{(z - z_\infty)(z_1 - z_0)}, \end{align*} (with appropriate modifications when one of $z_0, z_1, z_\infty = \infty$). One verifies $g(z_0) = 0$, $g(z_1) = 1$, $g(z_\infty) = \infty$, and $ad - bc = (z_0 - z_\infty)(z_1 - z_\infty)(z_1 - z_0) \neq 0$ (since the three points are distinct). Similarly, construct $h \in \mathcal{M}$ with $h(w_0) = 0$, $h(w_1) = 1$, $h(w_\infty) = \infty$. Then $f = h^{-1} \circ g \in \mathcal{M}$ satisfies $f(z_i) = w_i$ for $i = 0, 1, \infty$. **Step 2: Uniqueness.** [claim:Only Identity Fixes Zero One Infinity] If $\varphi \in \mathcal{M}$ satisfies $\varphi(0) = 0$, $\varphi(1) = 1$, and $\varphi(\infty) = \infty$, then $\varphi = \mathrm{id}$. [/claim] [proof] Write $\varphi(z) = \frac{az + b}{cz + d}$. From $\varphi(\infty) = \infty$: $c = 0$. From $\varphi(0) = 0$: $b = 0$. From $\varphi(1) = 1$: $a = d$. So $\varphi(z) = \frac{az}{a} = z$. [/proof] Suppose $f, f' \in \mathcal{M}$ both satisfy $f(z_i) = f'(z_i) = w_i$. Then $g \circ f^{-1} \circ f' \circ g^{-1}$ fixes $0, 1, \infty$, so it equals the identity by the claim above. Therefore $f^{-1} \circ f' = \mathrm{id}$, giving $f = f'$.