Left and Right Inverses of Mapping are Inverse Mapping/Proof 1

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Let $f: S \to T$ be a mapping such that:

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


$g_1 = g_2 = f^{-1}$

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


\(\ds g_2\) \(=\) \(\ds I_S \circ g_2\) Definition of Identity Mapping
\(\ds \) \(=\) \(\ds \paren {g_1 \circ f} \circ g_2\) by hypothesis
\(\ds \) \(=\) \(\ds g_1 \circ \paren {f \circ g_2}\) Composition of Mappings is Associative
\(\ds \) \(=\) \(\ds g_1 \circ I_T\) by hypothesis
\(\ds \) \(=\) \(\ds g_1\) Definition of Identity Mapping

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

It follows from Composite of Bijection with Inverse is Identity Mapping that $g_1 = g_2 = f^{-1}$.