Injection iff Left Cancellable/Necessary Condition

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Theorem

Let $f: Y \to Z$ be an injection.


Then $f$ is left cancellable.


Proof

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

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


Let $f: Y \to Z$ be an injection.

Let $X$ be a set

Let $g_1: X \to Y, g_2: X \to Y$ be mappings such that:

$f \circ g_1 = f \circ g_2$


Then $\forall x \in X$:

\(\ds \map f {g_1 \paren x}\) \(=\) \(\ds \map {f \circ g_1} x\) Definition of Composite Mapping
\(\ds \) \(=\) \(\ds \map {f \circ g_2} x\) By Hypothesis
\(\ds \) \(=\) \(\ds \map f {\map {g_2} x}\) Definition of Composite Mapping


As $f$ is an injection, $\map {g_1} x = \map {g_2} x$ and thus the condition for left cancellability holds.

$\blacksquare$


Sources