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.

Proof

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

• Inverse of Bijection

• Bijection Composite with Inverse for the converse of this result.

Sources

• George McCarty: Topology: An Introduction with Application to Topological Groups (1967)... (previous)... (next): $\text{I}$: Composition of Functions
• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.7$: Proposition $\text{A}.7.4$
• Paul Halmos and Steven Givant: Introduction to Boolean Algebras (2008)... (previous)... (next): Appendix $\text{A}$: Set Theory: Bijections