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 $x = \map {f^{-1} } y$.
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 $f^{-1} = g_1$.
Also:
\(\ds \map f x\) | \(=\) | \(\ds y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {I_T} y\) | $I_T$ 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 $f^{-1} = g_2$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.4$