The strategy for the forward direction is a direct computation showing cross-ratios are preserved. The backward direction constructs the Möbius map by routing both quadruples through $(0, 1, \infty)$.

**Step 1: Forward — Möbius maps preserve cross-ratios.**

Let $f(z) = \frac{az + b}{cz + d}$ with $ad - bc \neq 0$. For $z_j, z_k$ with $f(z_j), f(z_k) \neq \infty$ (i.e., $cz_j + d \neq 0$ and $cz_k + d \neq 0$): \begin{align*} f(z_j) - f(z_k) = \frac{(ad - bc)(z_j - z_k)}{(cz_j + d)(cz_k + d)}. \end{align*} The cross-ratio involves four such differences in the pattern $\frac{(w_1 - w_3)(w_2 - w_4)}{(w_1 - w_2)(w_3 - w_4)}$. Substituting $w_i = f(z_i)$, every factor of $(ad - bc)$ and $(cz_i + d)$ cancels in the ratio, giving $[f(z_1), f(z_2), f(z_3), f(z_4)] = [z_1, z_2, z_3, z_4]$. The special cases involving $\infty$ are verified by taking limits.

**Step 2: Backward — equal cross-ratios imply a Möbius map exists.**

Suppose $[z_1, z_2, z_3, z_4] = [w_1, w_2, w_3, w_4]$. By the [Sharp Triple Transitivity of Möbius Group](/theorems/811) theorem, there exist unique $g, h \in \mathcal{M}$ with: \begin{align*} g(z_1) = 0, \; g(z_2) = 1, \; g(z_4) = \infty, \qquad h(w_1) = 0, \; h(w_2) = 1, \; h(w_4) = \infty. \end{align*} By Step 1, $g(z_3) = [0, 1, g(z_3), \infty] = [g(z_1), g(z_2), g(z_3), g(z_4)] = [z_1, z_2, z_3, z_4]$. Similarly, $h(w_3) = [w_1, w_2, w_3, w_4]$. Since the cross-ratios are equal, $g(z_3) = h(w_3)$.

Set $f = h^{-1} \circ g$. Then $f(z_1) = w_1$, $f(z_2) = w_2$, $f(z_4) = w_4$, and $f(z_3) = h^{-1}(g(z_3)) = h^{-1}(h(w_3)) = w_3$.