Theorem: Möbius Transformation is Bijection

Let $a, b, c, d \in \C$ be complex numbers.
Let $f: \overline \C \to \overline \C$ be the Möbius transformation:
:<nowiki>$\map f z = \begin {cases} \dfrac {a z + b} {c z + d} & : z \ne -\dfrac d c \\
\infty & : z = -\dfrac d c \\
\dfrac a c & : z = \infty \\
\infty & : z = \infty \text { and } c = 0 \end {cases}$</nowiki>
where $\overline \C$ denotes the extended complex plane.
Then:
:$f: \overline \C \to \overline \C$ is a bijection
{{iff}}:
:$a c - b d \ne 0$

Proof:
We demonstrate that $f$ is injective {{iff}} $b c - a d \ne 0$.
{{begin-eqn}}
{{eqn | l = \map f {z_1}
      | r = \map f {z_2}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \dfrac {a z_1 + b} {c z_1 + d}
      | r = \dfrac {a z_2 + b} {c z_2 + d}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {a z_1 + b} \paren {c z_2 + d}
      | r = \paren {a z_2 + b} \paren {c z_1 + d}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = a c z_1 z_2 + b c z_2 + a d z_1 + b c
      | r = a c z_2 z_1 + b c z_1 + a d z_2 + b c
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = b c z_2 + a d z_1
      | r = b c z_1 + a d z_2
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {b c - a d} z_2
      | r = \paren {b c - a d} z_1
      | c = 
}}
{{end-eqn}}
demonstrating that when $z \ne -\dfrac d c$ and $z \ne \infty$:
:$\map f {z_1} = \map f {z_2} \implies z_1 = z_2$ {{iff}} $b c - a d \ne 0$
It remains to investigate the edge cases.
First we look at the case where $c \ne 0$.
{{begin-eqn}}
{{eqn | l = \dfrac {a z + b} {c z + d}
      | r = \dfrac a c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \paren {a z + b}
      | r = a \paren {c z + d}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a c z + b c
      | r = a c z + a d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b c - a d
      | r = 0
      | c = 
}}
{{end-eqn}}
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 $z \ne \infty$:
{{begin-eqn}}
{{eqn | l = w
      | r = \dfrac {a z + b} {c z + d}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = w \paren {c z + d}
      | r = a z + b
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = w c z + w d
      | r = a z + b
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = w c z - a z
      | r = b - w d
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = z \paren {c w - a}
      | r = - d w + b
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = z
      | r = \dfrac {- d w + b} {c w - a}
      | c = 
}}
{{end-eqn}}
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:
:$\map {f^{-1} } {w_1} = \map {f^{-1} } {w_2} \implies w_1 = w_2$ {{iff}} $\paren {-d} c - b \paren {-a} = 0$
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 {{iff}} $b c - a d \ne 0$.
It follows from Injection is Bijection iff Inverse is Injection that $f$ is a bijection.
{{qed}}
Category:Möbius Transformation is Bijection
Category:Möbius Transformations
Category:Examples of Bijections
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