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 Transformations ksdlnkksdje2sfvkw7hllf1ifjl61xs