Bijection iff Left and Right Inverse

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

$f$ is a bijection iff:


 * $\exists g_1: T \to S: g_1 \circ f = I_S$
 * $\exists g_2: T \to S: f \circ g_2 = I_T$

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

It also follows that it is necessarily the case that $g_1 = g_2$ for such to be possible.

Corollary
Let $f: S \to T$ and $g: T \to S$ be mappings such that:


 * $g \circ f = I_S$
 * $f \circ g = I_T$

Then both $f$ and $g$ are bijections.

Necessary Condition
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.

Sufficient Condition
Suppose $f$ is 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.

Now we need to show that $g_1 = g_2$.

Thus:

Proof of Corollary
Suppose we have such mappings $f$ and $g$ with the given properties.

From the main result, we have that $f$ is a bijection, by considering $g = g_1$ and $g = g_2$.

It directly follows that by setting $g = f, f = g_1, f = g_2$, the same result can be used the other way about.

Note: Axiom of Choice
This proof depends on the controversial Axiom of Choice (via Surjection iff Right Inverse).

Proof 2
Let $f: S \to T$ be a bijection.

From Bijection iff Inverse is Bijection and Bijection Composite with Inverse, it is shown that the inverse mapping $f^{-1}$ such that:
 * $f^{-1} \circ f = I_S$
 * $f \circ f^{-1} = I_T$

Hence we obtain the same result without recourse to the Axiom of Choice.

Proof 3
Follows directly from Inverse Relation is Left and Right Inverse iff Bijection.

Also see

 * Two-Sided Inverse