# 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
$a c - b d \ne 0$

### Restriction to Reals

Let $a, b, c, d \in \R$ be real numbers.

Let $f: \R^* \to \R^*$ be the Möbius transformation restricted to the real numbers:

$\map f x = \begin {cases} \dfrac {a x + b} {c x + d} & : x \ne -\dfrac d c \\ \infty & : x = -\dfrac d c \\ \dfrac a c & : x = \infty \\ \infty & : x = \infty \text { and } c = 0 \end {cases}$

Then:

$f: \R^* \to \R^*$ is a bijection
$a c - b d \ne 0$

## Proof

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

 $\ds \map f {z_1}$ $=$ $\ds \map f {z_2}$ $\ds \leadstoandfrom \ \$ $\ds \dfrac {a z_1 + b} {c z_1 + d}$ $=$ $\ds \dfrac {a z_2 + b} {c z_2 + d}$ $\ds \leadstoandfrom \ \$ $\ds \paren {a z_1 + b} \paren {c z_2 + d}$ $=$ $\ds \paren {a z_2 + b} \paren {c z_1 + d}$ $\ds \leadstoandfrom \ \$ $\ds a c z_1 z_2 + b c z_2 + a d z_1 + b c$ $=$ $\ds a c z_2 z_1 + b c z_1 + a d z_2 + b c$ $\ds \leadstoandfrom \ \$ $\ds b c z_2 + a d z_1$ $=$ $\ds b c z_1 + a d z_2$ $\ds \leadstoandfrom \ \$ $\ds \paren {b c - a d} z_2$ $=$ $\ds \paren {b c - a d} z_1$

demonstrating that when $z \ne -\dfrac d c$ and $z \ne \infty$:

$\map f {z_1} = \map f {z_2} \implies z_1 = z_2$ if and only if $b c - a d \ne 0$

It remains to investigate the edge cases.

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

 $\ds \dfrac {a z + b} {c z + d}$ $=$ $\ds \dfrac a c$ $\ds \leadsto \ \$ $\ds c \paren {a z + b}$ $=$ $\ds a \paren {c z + d}$ $\ds \leadsto \ \$ $\ds a c z + b c$ $=$ $\ds a c z + a d$ $\ds \leadsto \ \$ $\ds b c - a d$ $=$ $\ds 0$

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 $z \ne \infty$:

 $\ds w$ $=$ $\ds \dfrac {a z + b} {c z + d}$ $\ds \leadstoandfrom \ \$ $\ds w \paren {c z + d}$ $=$ $\ds a z + b$ $\ds \leadstoandfrom \ \$ $\ds w c z + w d$ $=$ $\ds a z + b$ $\ds \leadstoandfrom \ \$ $\ds w c z - a z$ $=$ $\ds b - w d$ $\ds \leadstoandfrom \ \$ $\ds z \paren {c w - a}$ $=$ $\ds - d w + b$ $\ds \leadstoandfrom \ \$ $\ds z$ $=$ $\ds \dfrac {- d w + b} {c w - a}$

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:

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

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 if and only if $b c - a d \ne 0$.

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

$\blacksquare$