Bijection has Left and Right Inverse/Proof 1

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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

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.

$\blacksquare$


Axiom of Choice

This proof depends on the Axiom of Choice, by way of Surjection iff Right Inverse.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.


Sources