Injection iff Left Cancellable/Sufficient Condition

Theorem
Let $f: Y \to Z$ be a mapping which is left cancellable.

Then $f$ is an injection.

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


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

Let $f: Y \to Z$ be left cancellable.

Aiming for a contradiction, suppose $f: Y \to Z$ were not injective.

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

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


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


 * $\forall y \in Y: g_2 \left({y}\right) = \begin{cases}

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

Thus we have $g_1 \ne g_2$ such that $f \circ g_1 = f \circ g_2$.

That is, $f$ is not left cancellable.

From this contradiction it follows that $f$ must be an injection after all.