# 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

 $\displaystyle g_2$ $=$ $\displaystyle I_S \circ g_2$ Definition of Identity Mapping $\displaystyle$ $=$ $\displaystyle \paren {g_1 \circ f} \circ g_2$ by hypothesis $\displaystyle$ $=$ $\displaystyle g_1 \circ \paren {f \circ g_2}$ Composition of Mappings is Associative $\displaystyle$ $=$ $\displaystyle g_1 \circ I_T$ by hypothesis $\displaystyle$ $=$ $\displaystyle g_1$ 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 $x = f^{-1} \left({y}\right)$.

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

Then $y = f \left({x}\right)$ and so:

 $\displaystyle f^{-1} \left({y}\right)$ $=$ $\displaystyle x$ $\displaystyle$ $=$ $\displaystyle I_S \left({x}\right)$ $I_S$ is the identity mapping on $S$ $\displaystyle$ $=$ $\displaystyle g_1 \circ f \left({x}\right)$ by hypothesis $\displaystyle$ $=$ $\displaystyle g_1 \left({y}\right)$ from above: $y = f \left({x}\right)$

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

Also:

 $\displaystyle f \left({x}\right)$ $=$ $\displaystyle y$ $\displaystyle$ $=$ $\displaystyle I_T \left({y}\right)$ $I_T$ is the identity mapping on $T$ $\displaystyle$ $=$ $\displaystyle f \circ g_2 \left({y}\right)$ by hypothesis $\displaystyle \implies \ \$ $\displaystyle f^{-1} \left({y}\right)$ $=$ $\displaystyle x$ as $f$ is injective $\displaystyle$ $=$ $\displaystyle g_2 \left({y}\right)$

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

$\blacksquare$

## Proof 3

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

 $(1):\quad$ $\displaystyle g_1 \circ f$ $=$ $\displaystyle I_S$ $\displaystyle \implies \ \$ $\displaystyle g_1 \circ f \circ g_2$ $=$ $\displaystyle I_S \circ g_2$ $\displaystyle$ $=$ $\displaystyle g_2$ Definition of Identity Mapping

 $(2):\quad$ $\displaystyle f \circ g_2$ $=$ $\displaystyle I_T$ $\displaystyle \implies \ \$ $\displaystyle g_1 \circ f \circ g_2$ $=$ $\displaystyle g_1 \circ I_T$ $\displaystyle$ $=$ $\displaystyle g_1$ 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 $f^{-1}$.

$\blacksquare$