[proofplan] We realize the Riemann sphere as the complex projective line, so that a nonsingular coefficient list acts on homogeneous coordinates by a [linear map](/page/Linear%20Map). The proposed inverse coefficient list is the adjugate list. Multiplying the original linear map and the adjugate linear map gives the nonzero scalar $(ad-bc)$ times the identity, which induces the identity map on projective coordinates. Finally, translating the adjugate action back to the affine chart gives the displayed formula for $T^{-1}$, including the point at infinity and denominator-zero cases. [/proofplan]

[step:Represent the Möbius transformation on homogeneous coordinates]Let $\Delta \in \mathbb{C}^{\times}$ denote the determinant \begin{align*} \Delta=ad-bc. \end{align*} Identify $\widehat{\mathbb{C}}$ with the complex projective line $\mathbb{P}^1(\mathbb{C})$ by sending a finite point $z \in \mathbb{C}$ to the homogeneous class $[z:1]$ and sending $\infty$ to $[1:0]$. Define the complex-linear map $A: \mathbb{C}^2 \to \mathbb{C}^2$ by \begin{align*} A(x,y)=(ax+by,cx+dy). \end{align*} Since $\Delta \ne 0$, the map $A$ is invertible, so $A(x,y) \ne (0,0)$ whenever $(x,y) \ne (0,0)$. Hence $A$ induces a well-defined map $\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}$ by \begin{align*} [x:y]\mapsto [ax+by:cx+dy]. \end{align*} On the affine chart $y \ne 0$, writing $z=x/y$, this induced map is exactly \begin{align*} z\mapsto \frac{az+b}{cz+d} \end{align*} with the usual convention that the value is $\infty$ when $cz+d=0$. Thus this induced projective map is $T$.[/step]

[guided]The clean way to control all extended values at once is to use homogeneous coordinates. We identify $\widehat{\mathbb{C}}$ with $\mathbb{P}^1(\mathbb{C})$: a finite complex number $z$ is represented by $[z:1]$, and the point $\infty$ is represented by $[1:0]$. Let $\Delta \in \mathbb{C}^{\times}$ be defined by \begin{align*} \Delta=ad-bc. \end{align*} The condition $\Delta \ne 0$ says exactly that the complex-linear map $A: \mathbb{C}^2 \to \mathbb{C}^2$ given by \begin{align*} A(x,y)=(ax+by,cx+dy) \end{align*} is invertible. Therefore $A$ never sends a nonzero vector of $\mathbb{C}^2$ to the zero vector. This is the needed well-definedness condition for projective space: if $[x:y]$ is a projective point, then \begin{align*} [x:y]\mapsto [ax+by:cx+dy] \end{align*} again gives a valid projective point. We also check that this projective formula is the [Möbius transformation](/page/M%C3%B6bius%20Transformation) stated in affine notation. If $y \ne 0$, define $z=x/y$. Then \begin{align*} [ax+by:cx+dy]=[az+b:cz+d]. \end{align*} If $cz+d \ne 0$, this is the finite point \begin{align*} \frac{az+b}{cz+d}. \end{align*} If $cz+d=0$, the second homogeneous coordinate is $0$, so the image is $[1:0]=\infty$. Thus the homogeneous-coordinate action is precisely the extended-value interpretation of $T$.[/guided]

[step:Compute the adjugate action and its compositions] Define the complex-linear map $B: \mathbb{C}^2 \to \mathbb{C}^2$ by \begin{align*} B(x,y)=(dx-by,-cx+ay). \end{align*} For every $(x,y) \in \mathbb{C}^2$, direct expansion gives \begin{align*} A(B(x,y))=(\Delta x,\Delta y). \end{align*} Indeed, the first coordinate is \begin{align*} a(dx-by)+b(-cx+ay)=(ad-bc)x, \end{align*} and the second coordinate is \begin{align*} c(dx-by)+d(-cx+ay)=(ad-bc)y. \end{align*} The same computation in the other order gives \begin{align*} B(A(x,y))=(\Delta x,\Delta y). \end{align*} Since multiplying both homogeneous coordinates by the nonzero scalar $\Delta$ does not change a point of $\mathbb{P}^1(\mathbb{C})$, the projective maps induced by $A$ and $B$ compose in both orders to the identity on $\widehat{\mathbb{C}}$. [/step]

[step:Translate the adjugate action into the inverse formula] Because the projective maps induced by $A$ and $B$ are two-sided inverses, the map induced by $B$ is $T^{-1}$. For a finite point $w \in \mathbb{C}$, represented by $[w:1]$, the map induced by $B$ sends \begin{align*} [w:1]\mapsto [dw-b:-cw+a]. \end{align*} Thus, when $-cw+a \ne 0$, \begin{align*} T^{-1}(w)=\frac{dw-b}{-cw+a}. \end{align*} When $-cw+a=0$, the image has second homogeneous coordinate $0$, so $T^{-1}(w)=\infty$. Finally, at $w=\infty$, represented by $[1:0]$, the inverse map sends \begin{align*} [1:0]\mapsto [d:-c]. \end{align*} This is the finite value $-d/c$ when $c \ne 0$, and it is $\infty$ when $c=0$. These are exactly the standard extended-value conventions for the formula \begin{align*} T^{-1}(w)=\frac{dw-b}{-cw+a}. \end{align*} Hence $T^{-1}$ is represented by the entries $d,-b,-c,a$, as claimed. [/step]