# Left and Right Inverse Mappings Implies Bijection

## Theorem

Let $f: S \to T$ be a mapping.

Let $f$ be such that:

- $\exists g_1: T \to S: g_1 \circ f = I_S$
- $\exists g_2: T \to S: f \circ g_2 = I_T$

where both $g_1$ and $g_2$ are mappings.

Then $f$ is a bijection.

## Proof 1

Suppose:

- $\exists g_1: T \to S: g_1 \circ f = I_S$
- $\exists g_2: T \to S: f \circ g_2 = I_T$

We have that the Identity Mapping is Bijection, thus $I_S$ and $I_T$ are both bijections.

From Injection if Composite is Injection, if $I_S = g_1 \circ f$ is an injection, then so is $f$.

From Surjection if Composite is Surjection, if $I_T = f \circ g_2$ is a surjection, then so is $f$.

So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.

$\blacksquare$

## Proof 2

Suppose:

- $\exists g_1: T \to S: g_1 \circ f = I_S$
- $\exists g_2: T \to S: f \circ g_2 = I_T$

From Injection iff Left Inverse, it follows that $f$ is an injection.

From Surjection iff Right Inverse, it follows that $f$ is a surjection.

So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $1.3: \ 11$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $4$: Mappings: Exercise $15$