# Quotient Mapping is Surjection

## Theorem

Let $\mathcal R$ be an equivalence relation on $S$.

Then the quotient mapping $q_{\mathcal R}: S \to S / \mathcal R$ is a surjection.

## Proof

From Equivalence Class is not Empty, we have that:

$\forall \eqclass x {\mathcal R} \in S / \mathcal R: \exists x \in S: x \in \eqclass x {\mathcal R}$

and the result follows.

$\blacksquare$