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
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $14$