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.

$\blacksquare$