# 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 $f^{-1}$.

$\blacksquare$