# Surjection iff Right Cancellable/Necessary Condition/Proof 1

## Theorem

Let $f$ be a surjection.

Then $f$ is right cancellable.

## Proof

Let $f: X \to Y$ be surjective.

Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$.

As $f$ is a surjection:

- $\Img f = Y$

by definition.

But in order for $h_1 \circ f$ to be defined, it is necessary that $Y = \Dom {h_1}$.

Similarly, for $h_2 \circ f$ to be defined, it is necessary that $Y = \Dom {h_2}$.

So it follows that the domains of $h_1$ and $h_2$ are the same.

Also:

- The codomain of $h_1$ equals the codomain of $h_1 \circ f$
- The codomain of $h_2$ equals the codomain of $h_2 \circ f$

again by definition of composition of mappings.

Now, we have shown that the domains and codomains of $h_1$ and $h_2$ are the same.

All we need to do now to prove that $h_1 = h_2$, and therefore that $f$ is right cancellable, is to show that:

- $\forall y \in Y: h_1 \paren y = h_2 \paren y$.

So, let $y \in Y$.

As $f$ is surjective:

- $\exists x \in X: y = f \paren x$

Thus:

\(\ds h_1 \paren y\) | \(=\) | \(\ds h_1 \paren {f \paren x}\) | Definition of $y$ | |||||||||||

\(\ds \) | \(=\) | \(\ds h_1 \circ f \paren x\) | Definition of Composition of Mappings | |||||||||||

\(\ds \) | \(=\) | \(\ds h_2 \circ f \paren x\) | By Hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds h_2 \paren {f \paren x}\) | Definition of Composition of Mappings | |||||||||||

\(\ds \) | \(=\) | \(\ds h_2 \paren y\) | Definition of $y$ |

Thus $h_1 \paren y = h_2 \paren y$ and thus $f$ is right cancellable.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 25$: Some further results and examples on mappings: Worked Example $1$