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

Jump to navigation
Jump to search

## 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 $15$