Left and Right Inverses of Mapping are Inverse Mapping/Proof 3

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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

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 $f^{-1}$.

$\blacksquare$


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