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

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Let $f: S \to T$ be a mapping such that:

$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$


$g_1 = g_2 = f^{-1}$

where $f^{-1}$ is the inverse of $f$.


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 $f^{-1} = g_1$.


\(\ds \map f x\) \(=\) \(\ds y\)
\(\ds \) \(=\) \(\ds \map {I_T} y\) $I_T$ 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 $f^{-1} = g_2$.