Trivial Quotient is a Bijection

Theorem
Let $\Delta_S$ be the diagonal relation on a set $S$.

Let $q_{\Delta_S}: S \to S / \Delta_S$ be the trivial quotient of $S$.

Then $q_{\Delta_S}: S \to S / \Delta_S$ is a bijection.

Proof
The diagonal relation is defined as:


 * $\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$

From the fact that $q_{\Delta_S}$ is a quotient mapping, we know that it is a surjection.

It relates each $x \in S$ to the singleton $\set x$.

Thus:
 * $\set x = \eqclass x {\Delta_S} \subseteq S$

So $\map {q_{\Delta_S} } x = \set x$, and it follows that:


 * $\eqclass x {\Delta_S} = \eqclass y {\Delta_S} \implies x = y$

Thus $q_{\Delta_S}$ is injective, and therefore by definition a bijection.

Comment
Some sources abuse notation and write $\map {q_{\Delta_S} } x = x$, which serves to emphasise its triviality, but can cause it to be conflated with the identity mapping.