# 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 $5.4$