Möbius Transformation is Bijection/Restriction to Reals

Theorem
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


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

Proof
First we note that as Real Addition is Closed and Real Multiplication is Closed:
 * $\Dom {\R^*} \subseteq \R^*$

Recall from Möbius Transformation is Bijection that the Möbius transformation on the extended complex plane is a bijection $a c - b d \ne 0$.

From Restriction of Injection is Injection, if $a c - b d \ne 0$ $f$ is an injection.

As the inverse $f^{-1}$ of $f$ is also the restriction of a Möbius transformation, it follows that $f^{-1}$ is also an injection.

Hence the result from Injection is Bijection iff Inverse is Injection.