Composition of Right Inverse with Mapping is Idempotent

Theorem

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

Let $g: T \to S$ be a right inverse mapping of $f$.

Then:

$\paren {g \circ f} \circ \paren {g \circ f} = g \circ f$

Proof

\(\ds \paren {g \circ f} \circ \paren {g \circ f}\)

\(=\)

\(\ds g \circ \paren {f \circ g} \circ f\)

Composition of Mappings is Associative

\(\ds \)

\(=\)

\(\ds g \circ I_T \circ f\)

Definition of Right Inverse Mapping

\(\ds \)

\(=\)

\(\ds g \circ f\)

Definition of Identity Mapping

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $3$. Mappings: Exercise $6$