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

$f \circ f^{-1} = I_T$

$\blacksquare$