Bijection iff exists Mapping which is Left and Right Inverse
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then $f$ is a bijection if and only if:
- there exists a mapping $g: T \to S$ such that:
- $g \circ f = I_S$
- $f \circ g = I_T$
- where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.
Proof
Necessary Condition
Let $f$ be a bijection.
Then for each $y \in T$ there exists one and only one $x \in S$ such that $\map f x = y$.
That is, that there exists a mapping $g: T \to S$ with the property that:
- $\forall y \in T: \exists x \in S: \map g y = x$
Let $y \in T$.
Let $x = g \map g y$.
Then:
\(\ds y\) | \(=\) | \(\ds \map f {\map g y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {f \circ g} y\) |
and:
\(\ds x\) | \(=\) | \(\ds \map g {\map f x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {f \circ g} x\) |
demonstrating that $g$ has the property that:
- $g \circ f = I_S$
- $f \circ g = I_T$
$\Box$
Sufficient Condition
Suppose there exists a $g$ which satisfies the conditions on $f$.
By Condition for Composite Mapping to be Identity:
- $(1): \quad$ from $g \circ f = I_S$ it follows that $f$ is an injection.
- $(2): \quad$ from $f \circ g = I_T$ it follows that $f$ is a surjection.
Thus, by definition, $f$ is a bijection.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Theorem $2$