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.

$\blacksquare$

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.

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Quotient Functions