Bijection has Left and Right Inverse/Proof 2
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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
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.
$\blacksquare$