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