Left and Right Inverse Mappings Implies Bijection/Proof 2

Theorem

Let $f: S \to T$ be a mapping.

Let $f$ be such that:

$\exists g_1: T \to S: g_1 \circ f = I_S$

$\exists g_2: T \to S: f \circ g_2 = I_T$

where both $g_1$ and $g_2$ are mappings.

Then $f$ is a bijection.

Proof

Suppose:

$\exists g_1: T \to S: g_1 \circ f = I_S$

$\exists g_2: T \to S: f \circ g_2 = I_T$

From Injection iff Left Inverse, it follows that $f$ is an injection.

From Surjection iff Right Inverse, it follows that $f$ is a surjection.

So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.

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.