Left and Right Inverses of Mapping are Inverse Mapping/Proof 2

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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

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 $f^{-1} = g_2$.

$\blacksquare$


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