Proof By this characterization, **T**: **X** → **Y** is onto means there is **S**: **Y** → **X** satisfying **TS** = `id`_{Y}. By this characterization, **T**: **X** → **Y** is one-to-one means there is **S'**: **Y** → **X** satisfying **S'T** = `id`_{X}. Thus

**T**: **X** → **Y** is onto and one-to-one ⇔ There are **S**, **S'**: **Y** → **X** satisfying **TS** = `id`_{Y} and **S'T** = `id`_{X}.

The proof is then complete by showing that **S** and **S'** must be equal:

**TS** = `id`_{Y} and **S'T** = `id`_{X} ⇒ **S** = `id`_{X}S = **S'TS** = `S'id`_{Y} = `S'`.