Injection iff Left Cancellable

Theorem
A mapping $f$ is an injection iff $f$ is left cancellable.

Proof
From the definition: a mapping $f: Y \to Z$ is left cancellable if:


 * $\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$


 * Suppose $f: Y \to Z$ is injective.

Suppose $g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2$.

Then $\forall x \in X$:

Thus as $f$ is an injection, $g_1 \left({x}\right) = g_2 \left({x}\right)$ and thus the condition for left cancellability holds.


 * Suppose $f: Y \to Z$ is not injective.

Then $\exists y_1 \ne y_2 \in Y: f \left({y_1}\right) = f \left({y_2}\right)$.

For $f$ to be left cancellable, the condition $f \circ g_1 = f \circ g_2 \implies g_1 = g_2$ needs to apply to all mappings $g_1$ and $g_2$, whatever they are. So if we can always create two mappings $g_1$ and $g_2$ such that this condition does not hold, we have proved the supposition.

So, let us create these mappings on the domain $X = Y$.

Let the two mappings $g_1: X \to Y, g_2: X \to Y$ be defined as follows:


 * $g_1 \left({y}\right) = y: y \in X$


 * $g_2 \left({y}\right) = \begin{cases}

y_2 & : y = y_1 \\ y & : y \ne y_1 \end{cases}$

So, for an arbitrary non-injective mapping $f$, we have created two unequal functions $g_1$ and $g_2$ both of which fulfil the conditions, and therefore $f$ is not left cancellable.

The result follows.

Also see

 * Surjection iff Right Cancellable