Möbius Transformation is Bijection

Theorem
Let $a, b, c, d \in \C$ be complex numbers.

Let $f: \overline \C \to \overline \C$ be the Möbius transformation:


 * $\map f z = \begin {cases} \dfrac {a z + b} {c z + d} & : z \ne -\dfrac d c \\

\infty & : z = -\dfrac d c \\ \dfrac a c & : z = \infty \\ \infty & : z = \infty \text { and } c = 0 \end {cases}$

Then:
 * $f: \overline \C \to \overline \C$ is a bijection


 * $a c - b d \ne 0$
 * $a c - b d \ne 0$

Proof
We demonstrate that $f$ is injective $b c - a d \ne 0$.

demonstrating that when $z \ne -\dfrac d c$ and $z \ne \infty$:
 * $\map f {z_1} = \map f {z_2} \implies z_1 = z_2$ $b c - a d \ne 0$

It remains to investigate the edge cases.

First we look at the case where $c \ne 0$.

That is, for $z \in \R$:
 * $\dfrac {a z + b} {c z + d} = \dfrac c a$ only if $b c - a d = 0$

and so if $\map f {z_1} = \map f {z_2} = \dfrac a c$ it follows that $z_1 = z_2 = \infty$.

The case where $\map f {z_1} = \map f {z_2} = \infty$ follows by definition either that:
 * $z_1 = z_2 = \dfrac a c$ when $c \ne 0$

or:
 * $z_1 = z_2 = \infty$ when $c = 0$.

Thus we have that $f$ is an injection.

Now we investigate the inverse of $f$.

From Inverse Element of Injection we have that:


 * $\map f z = w \implies \map {f^{-1} } w = z$

So, let $w = \map f z$.

First we recall that if $z = -\dfrac d c$, then $c z + d = 0$ and so $\dfrac {a z + b} {c z + d}$ is undefined.

Hence the need to investigate that case separately.

Take the general case, where $z \ne -\dfrac d c$ and $z \ne \infty$:

Thus we have that:
 * $\map {f^{-1} } w = \dfrac {- d w + b} {c w - a}$

which is again a Möbius transformation, defined over all $w \in \C$ except where $w = \dfrac a c$.

We define:
 * $\map {f^{-1} } {\dfrac a c} = \infty$

and:
 * $\map {f^{-1} } \infty = -\dfrac d c$

except when $c = \infty$, where we define:
 * $\map {f^{-1} } \infty = \infty$

Hence we have that the inverse of $f$ is another Möbius transformation.

So as $f^{-1}$ is also a Möbius transformation, it follows that:


 * $\map {f^{-1} } {w_1} = \map {f^{-1} } {w_2} \implies w_1 = w_2$ $\paren {-d} c - b \paren {-a} = 0$

which is the same thing as $b c - a d \ne 0$.

Again, we have that $\dfrac {- d w + b} {c w - a} = -\dfrac d c$ only if $\paren {-d} c - b \paren {-a} = 0$.

As seen above, this is the same thing as $b c - a d \ne 0$.

Finally, we note that:
 * $\map {f^{-1} } {w_1} = \map {f^{-1} } {w_2} = \infty \implies w_1 = w_2 = \dfrac {-d} c$

Thus we have that $f^{-1}$ is injective $b c - a d \ne 0$.

It follows from Injection is Bijection iff Inverse is Injection that $f$ is a bijection.