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

Theorem

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

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

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

Then:

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

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

Proof

\(\ds g_2\)

\(=\)

\(\ds I_S \circ g_2\)

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

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.9$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $15$