[proofplan] We first build, for any ordered triple of pairwise distinct points in $\widehat{\mathbb{C}}$, an explicit [Möbius transformation](/page/M%C3%B6bius%20Transformation) sending that triple to $(0,1,\infty)$. Applying this construction to the source triple and the target triple gives two normalizing maps, and composing the inverse of the target normalizing map with the source one gives the required interpolation map. For uniqueness, we reduce to the special case of a Möbius transformation fixing $0$, $1$, and $\infty$, where comparing coefficients forces the transformation to be the identity. [/proofplan]

[step:Construct a normalizing Möbius transformation for any ordered triple]Let $(a,b,c)\in\widehat{\mathbb{C}}^3$ be an ordered triple of pairwise distinct points. We define a map \begin{align*} N_{a,b,c}:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} by the following four formulas, according to which entry is $\infty$. If $a,b,c\in\mathbb{C}$, define \begin{align*} N_{a,b,c}(z)=\frac{(b-c)(z-a)}{(b-a)(z-c)}. \end{align*} The determinant of the representing matrix is \begin{align*} (b-c)(b-a)(a-c), \end{align*} which is nonzero because $a,b,c$ are pairwise distinct. Therefore this is a Möbius transformation. Its extended values satisfy \begin{align*} N_{a,b,c}(a)=0,\qquad N_{a,b,c}(b)=1,\qquad N_{a,b,c}(c)=\infty. \end{align*} If $c=\infty$, then $a,b\in\mathbb{C}$ and $a\ne b$. Define \begin{align*} N_{a,b,\infty}(z)=\frac{z-a}{b-a}. \end{align*} This is represented by a matrix with determinant $1/(b-a)\ne 0$, so it is a Möbius transformation, and \begin{align*} N_{a,b,\infty}(a)=0,\qquad N_{a,b,\infty}(b)=1,\qquad N_{a,b,\infty}(\infty)=\infty. \end{align*} If $a=\infty$, then $b,c\in\mathbb{C}$ and $b\ne c$. Define \begin{align*} N_{\infty,b,c}(z)=\frac{b-c}{z-c}. \end{align*} This is represented by a matrix with determinant $-(b-c)\ne 0$, so it is a Möbius transformation, and \begin{align*} N_{\infty,b,c}(\infty)=0,\qquad N_{\infty,b,c}(b)=1,\qquad N_{\infty,b,c}(c)=\infty. \end{align*} If $b=\infty$, then $a,c\in\mathbb{C}$ and $a\ne c$. Define \begin{align*} N_{a,\infty,c}(z)=\frac{z-a}{z-c}. \end{align*} This is represented by a matrix with determinant $a-c\ne 0$, so it is a Möbius transformation, and \begin{align*} N_{a,\infty,c}(a)=0,\qquad N_{a,\infty,c}(\infty)=1,\qquad N_{a,\infty,c}(c)=\infty. \end{align*} Thus, in every case, $N_{a,b,c}$ is a Möbius transformation satisfying \begin{align*} N_{a,b,c}(a)=0,\qquad N_{a,b,c}(b)=1,\qquad N_{a,b,c}(c)=\infty. \end{align*}[/step]

[guided]The point of this step is to avoid hiding the cases involving $\infty$. We want one normalizing Möbius transformation for every ordered triple $(a,b,c)$ of pairwise distinct points, sending the first point to $0$, the second to $1$, and the third to $\infty$. When all three points are finite, the standard cross-ratio normalization is \begin{align*} N_{a,b,c}(z)=\frac{(b-c)(z-a)}{(b-a)(z-c)}. \end{align*} This is a fractional [linear map](/page/Linear%20Map). Its representing matrix has first row $((b-c),-(b-c)a)$ and second row $((b-a),-(b-a)c)$. The determinant is \begin{align*} (b-c)(b-a)(a-c). \end{align*} Since $a,b,c$ are pairwise distinct, each factor is nonzero, so the determinant is nonzero. Hence the formula defines a Möbius transformation. Substituting the three special values gives \begin{align*} N_{a,b,c}(a)=0, \end{align*} because the numerator vanishes, \begin{align*} N_{a,b,c}(b)=\frac{(b-c)(b-a)}{(b-a)(b-c)}=1, \end{align*} and \begin{align*} N_{a,b,c}(c)=\infty, \end{align*} because the denominator vanishes while the numerator does not. It remains to make the same normalization explicit when one of the points is $\infty$. If $c=\infty$, then $a,b\in\mathbb{C}$ and $a\ne b$, and the affine map \begin{align*} N_{a,b,\infty}(z)=\frac{z-a}{b-a} \end{align*} sends $a$ to $0$, sends $b$ to $1$, and sends $\infty$ to $\infty$. Its determinant is $1/(b-a)\ne 0$. If $a=\infty$, then $b,c\in\mathbb{C}$ and $b\ne c$. The map \begin{align*} N_{\infty,b,c}(z)=\frac{b-c}{z-c} \end{align*} has value $0$ at $\infty$, value $1$ at $b$, and value $\infty$ at $c$. Its determinant is $-(b-c)\ne 0$, so it is again Möbius. If $b=\infty$, then $a,c\in\mathbb{C}$ and $a\ne c$. The map \begin{align*} N_{a,\infty,c}(z)=\frac{z-a}{z-c} \end{align*} has value $0$ at $a$, value $1$ at $\infty$, and value $\infty$ at $c$. Its determinant is $a-c\ne 0$. These four cases exhaust all possibilities, since the triple is pairwise distinct and hence at most one of $a,b,c$ can be $\infty$.[/guided]

[step:Compose source and target normalizations to obtain the required map] Define \begin{align*} A:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} by \begin{align*} A=N_{z_1,z_2,z_3}. \end{align*} Define \begin{align*} B:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} by \begin{align*} B=N_{w_1,w_2,w_3}. \end{align*} By the previous step, $A$ and $B$ are Möbius transformations and satisfy \begin{align*} A(z_1)=0,\qquad A(z_2)=1,\qquad A(z_3)=\infty, \end{align*} and \begin{align*} B(w_1)=0,\qquad B(w_2)=1,\qquad B(w_3)=\infty. \end{align*} By [citetheorem:7865], $B$ is bijective and its inverse $B^{-1}:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}}$ is a Möbius transformation. By [citetheorem:7864], the composition of two Möbius transformations is represented by the product of their representing matrices, whose determinant is the product of the two nonzero determinants. Hence the map \begin{align*} T:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} defined by \begin{align*} T=B^{-1}\circ A \end{align*} is a Möbius transformation. For each $i\in\{1,2,3\}$, the equality $A(z_i)=B(w_i)$ gives \begin{align*} T(z_i)=B^{-1}(A(z_i))=B^{-1}(B(w_i))=w_i. \end{align*} Thus a Möbius transformation with the required three values exists. [/step]

[step:Show that a Möbius transformation fixing $0$, $1$, and $\infty$ is the identity] Let $\operatorname{id}_{\widehat{\mathbb{C}}}:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}}$ denote the identity map on $\widehat{\mathbb{C}}$, defined by $\operatorname{id}_{\widehat{\mathbb{C}}}(z)=z$ for every $z\in\widehat{\mathbb{C}}$. Let \begin{align*} M:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} be a Möbius transformation satisfying \begin{align*} M(0)=0,\qquad M(1)=1,\qquad M(\infty)=\infty. \end{align*} Choose a representative \begin{align*} M(z)=\frac{\alpha z+\beta}{\gamma z+\delta} \end{align*} with $\alpha,\beta,\gamma,\delta\in\mathbb{C}$ and $\alpha\delta-\beta\gamma\ne 0$. The condition $M(\infty)=\infty$ means $\gamma=0$. Since the determinant is nonzero, $\delta\ne 0$. The condition $M(0)=0$ gives \begin{align*} 0=M(0)=\frac{\beta}{\delta}, \end{align*} so $\beta=0$. The condition $M(1)=1$ then gives \begin{align*} 1=M(1)=\frac{\alpha}{\delta}, \end{align*} so $\alpha=\delta$. Therefore, for every finite $z\in\mathbb{C}$, \begin{align*} M(z)=\frac{\delta z}{\delta}=z, \end{align*} and also $M(\infty)=\infty$. Hence $M=\operatorname{id}_{\widehat{\mathbb{C}}}$. [/step]

[step:Conjugate two interpolating transformations to finish uniqueness] Let \begin{align*} S:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} be any Möbius transformation satisfying \begin{align*} S(z_i)=w_i \quad \text{for every } i\in\{1,2,3\}. \end{align*} Using the map $T$ constructed above, define \begin{align*} M:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}} \end{align*} by \begin{align*} M=A\circ S^{-1}\circ T\circ A^{-1}. \end{align*} By [citetheorem:7865], $S^{-1}:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}}$ and $A^{-1}:\widehat{\mathbb{C}}\to\widehat{\mathbb{C}}$ are Möbius transformations. By [citetheorem:7864], $M$ is a Möbius transformation because it is a composition of Möbius transformations. For each $i\in\{1,2,3\}$, we have $T(z_i)=w_i$ and $S(z_i)=w_i$, so $S^{-1}(w_i)=z_i$. Hence \begin{align*} M(A(z_i))=A(S^{-1}(T(z_i)))=A(S^{-1}(w_i))=A(z_i). \end{align*} Since \begin{align*} A(z_1)=0,\qquad A(z_2)=1,\qquad A(z_3)=\infty, \end{align*} the transformation $M$ fixes $0$, $1$, and $\infty$. By the previous step, \begin{align*} M=\operatorname{id}_{\widehat{\mathbb{C}}}. \end{align*} Thus \begin{align*} A\circ S^{-1}\circ T\circ A^{-1}=\operatorname{id}_{\widehat{\mathbb{C}}}. \end{align*} Composing on the left by $A^{-1}$ and on the right by $A$ gives \begin{align*} S^{-1}\circ T=\operatorname{id}_{\widehat{\mathbb{C}}}, \end{align*} and therefore $S=T$. The interpolating Möbius transformation is unique. [/step]