Absolute Value is Many-to-One

Theorem
Let $f: \R \to \R$ be the absolute value function:
 * $\forall x \in \R: \map f x = \begin{cases}

x & : x \ge 0 \\ -x & : x < 0 \end{cases}$

Then $f$ is a many-to-one relation.

Proof
$f$ is not a many-to-one relation.

Then there exists $y_1 \in \R$ and $y_2 \in \R$ where $y_1 \ne y_2$ such that:
 * $\exists x \in \R: \map f x = y_1, \map f x = y_2$

There are two possibilities:

Suppose $x \ge 0$.

Then:

That is:
 * $y_2 = y_1 = x$

Suppose $x < 0$.

Then:

That is:
 * $y_2 = y_1 = -x$

So by Proof by Cases we have that $y_1 = y_2$ whatever $x$ may be.

This contradicts our assertion that $y_1 \ne y_2$.

Hence it follows by Proof by Contradiction that $f$ is many-to-one.