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.