Real Square Function is not Injective

Example of Mapping which is Not an Injection
Let $f: \R \to R$ be the real function defined as:
 * $\forall x \in \R: f \paren x = x^2$

Then $f$ is not an injection.

Proof
For $f$ to be an injection, it would be necessary that:
 * $\forall x_1, x_2 \in \R: f \paren {x_1} = f \paren {x_2} \implies x_1 \ne x_2 \implies f \paren {x_1} \ne f \paren {x_2}$

By definition of the squaring operation, we have:
 * $f \paren x = f \paren {-x}$

But unless $x = 0$ it is not the case that $x = -x$.

Hence $f$ is not an injection.