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:
- $\map f z = \begin {cases} \dfrac {a z + b} {c z + d} & : z \ne -\dfrac d c \\ \infty & : z = -\dfrac d c \\ \dfrac a c & : z = \infty \\ \infty & : z = \infty \text { and } c = 0 \end {cases}$
where $\overline \C$ denotes the extended complex plane.
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.
$\blacksquare$