# 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 $f^{-1} = g_2$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 5$: Theorem $5.4$