# 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

- $a c - b d \ne 0$

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## Proof

First we note that as Real Addition is Closed and Real Multiplication is Closed:

- $\Dom {\R^*} \subseteq \R^*$

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

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

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions: Exercise $2$