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:

$\Delta_S = \left\{{\left({x, x}\right): x \in S}\right\} \subseteq S \times S$

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 $\left\{{x}\right\}$.

Thus $\left\{{x}\right\} = \left[\!\left[{x}\right]\!\right]_{\Delta_S} \subseteq S$.

So $q_{\Delta_S} \left({x}\right) = \left\{{x}\right\}$, and it follows that:
 * $\left[\!\left[{x}\right]\!\right]_{\Delta_S} = \left[\!\left[{y}\right]\!\right]_{\Delta_S} \implies x = y$

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

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