Composition of Right Inverse with Mapping is Idempotent

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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

\(\displaystyle \paren {g \circ f} \circ \paren {g \circ f}\) \(=\) \(\displaystyle g \circ \paren {f \circ g} \circ f\) $\quad$ Composition of Mappings is Associative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g \circ I_T \circ f\) $\quad$ Definition of Right Inverse Mapping $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g \circ f\) $\quad$ Definition of Identity Mapping $\quad$

$\blacksquare$


Sources