[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 $T$.
[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]
[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]
