Left and Right Inverses of Mapping are Inverse Mapping

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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

\(\displaystyle g_2\) \(=\) \(\displaystyle I_S \circ g_2\) $\quad$ Definition of Identity Mapping $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {g_1 \circ f} \circ g_2\) $\quad$ by hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1 \circ \paren {f \circ g_2}\) $\quad$ Composition of Mappings is Associative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1 \circ I_T\) $\quad$ by hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1\) $\quad$ Definition of Identity Mapping $\quad$


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:

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

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


Also:

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

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

$\blacksquare$


Proof 3

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

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


\((2):\quad\) \(\displaystyle f \circ g_2\) \(=\) \(\displaystyle I_T\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle g_1 \circ f \circ g_2\) \(=\) \(\displaystyle g_1 \circ I_T\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g_1\) $\quad$ Definition of Identity Mapping $\quad$

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 $f^{-1}$.

$\blacksquare$