# Left and Right Inverses of Mapping are Inverse Mapping

## 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 1

\(\ds g_2\) | \(=\) | \(\ds I_S \circ g_2\) | 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 $g_1 = g_2 = f^{-1}$.

$\blacksquare$

## Proof 2

From Left and Right Inverse Mappings Implies Bijection, we have that $f$ is a bijection.

Let $y \in T$ and let $x = \map {f^{-1} } y$.

Such an $x$ exists because $f$ is surjective, and unique within $S$ as $f$ is injective.

Then $y = \map f x$ and so:

\(\ds \map {f^{-1} } y\) | \(=\) | \(\ds x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map {I_S} x\) | $I_S$ is the identity mapping on $S$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {g_1 \circ f} x\) | by hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {g_1} y\) | from above: $y = \map f x$ |

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

Also:

\(\ds \map f x\) | \(=\) | \(\ds y\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map {I_T} y\) | $I_T$ is the identity mapping on $T$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {f \circ g_2} y\) | by hypothesis | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map {f^{-1} } y\) | \(=\) | \(\ds x\) | as $f$ is injective | ||||||||||

\(\ds \) | \(=\) | \(\ds \map {g_2} y\) |

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

$\blacksquare$

## Proof 3

Because Composition of Mappings is Associative, brackets do not need to be used.

\(\text {(1)}: \quad\) | \(\ds g_1 \circ f\) | \(=\) | \(\ds I_S\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds g_1 \circ f \circ g_2\) | \(=\) | \(\ds I_S \circ g_2\) | |||||||||||

\(\ds \) | \(=\) | \(\ds g_2\) | Definition of Identity Mapping |

\(\text {(2)}: \quad\) | \(\ds f \circ g_2\) | \(=\) | \(\ds I_T\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds g_1 \circ f \circ g_2\) | \(=\) | \(\ds g_1 \circ I_T\) | |||||||||||

\(\ds \) | \(=\) | \(\ds g_1\) | Definition of Identity Mapping |

Thus $g_1 = g_2$.

Now suppose there exists $g_3: T \to S: g_3 \circ f = I_S$.

By the same argument as above, $g_3 = g_2$.

This means that $g_1 (= g_3)$ is the only left inverse of $f$.

Similarly, suppose there exists $g_4: T \to S: f \circ g_4 = I_T$.

By the same argument as above, $g_4 = g_1$.

This means that $g_2 (= g_4)$ is the only right inverse of $f$.

So $g_1 = g_2 = g_3 = g_4$ are all the same.

By Composite of Bijection with Inverse is Identity Mapping, it follows that this unique mapping is the inverse $f^{-1}$.

$\blacksquare$