Real Square Function is not Injective

Example of Mapping which is Not an Injection

Let $f: \R \to \R$ be the real square function:

$\forall x \in \R: \map f x = x^2$

Then $f$ is not an injection.

Proof

For $f$ to be an injection, it would be necessary that:

$\forall x_1, x_2 \in \R: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$

By definition of the squaring operation, we have:

$\map f x = \map f {-x}$

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

Hence $f$ is not an injection.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 22$: Injections; bijections; inverse of a bijection