Bijection has Left and Right Inverse
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Theorem
Let $f: S \to T$ be a bijection.
Let:
- $I_S$ be the identity mapping on $S$
- $I_T$ be the identity mapping on $T$.
Let $f^{-1}$ be the inverse of $f$.
Then:
- $f^{-1} \circ f = I_S$
and:
- $f \circ f^{-1} = I_T$
where $\circ$ denotes composition of mappings.
Proof 1
Let $f$ be a bijection.
Then it is both an injection and a surjection, thus both the described $g_1$ and $g_2$ must exist from Injection iff Left Inverse and Surjection iff Right Inverse.
The fact that $g_1 = g_2 = f^{-1}$ follows from Left and Right Inverses of Mapping are Inverse Mapping.
$\blacksquare$
Proof 2
Suppose $f$ is a bijection.
From Bijection iff Inverse is Bijection and Composite of Bijection with Inverse is Identity Mapping, it is shown that the inverse mapping $f^{-1}$ such that:
- $f^{-1} \circ f = I_S$
- $f \circ f^{-1} = I_T$
is a bijection.
$\blacksquare$
Proof 3
Let $f$ be a bijection.
By definition, $f$ is a mapping, and hence also by definition a relation.
Hence the result Bijective Relation has Left and Right Inverse applies directly and so:
- $f^{-1} \circ f = I_S$
and
- $f \circ f^{-1} = I_T$
$\blacksquare$