Bijection iff Left and Right Inverse

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

$f$ is a bijection :


 * $(1)\quad$ $\exists g_1: T \to S: g_1 \circ f = I_S$
 * $(2)\quad$ $\exists g_2: T \to S: f \circ g_2 = I_T$

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.

Necessary Condition
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 from Left and Right Inverse Mappings Implies Bijection it follows that $f$ is a bijection.

From Left and Right Inverses of Mapping are Inverse Mapping it follows that:
 * $g_1 = g_2 = f^{-1}$

where $f^{-1}$ is the inverse of $f$.

Sufficient Condition
Let $f: S \to T$ be a bijection.

Then from Bijection has Left and Right Inverse it follows that:
 * $f^{-1} \circ f = I_S$ and
 * $f \circ f^{-1} = I_T$

where $f^{-1}$ is the inverse of $f$.

Also see

 * Composite of Bijection with Inverse is Identity Mapping for the converse of this result.