Diagonal Relation is Right Identity

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$.

Then:
 * $\mathcal R \circ \Delta_S = \mathcal R$

where $\Delta_S$ is the diagonal relation on $S$, and $\circ$ signifies composition of relations.

Proof
We use the definition of relation equality, as follows:

Equality of Codomains
The codomains of $\mathcal R$ and $\mathcal R \circ \Delta_S$ are both equal to $T$ from Codomain of Composite Relation.

Equality of Domains
The domains of $\mathcal R$ and $\mathcal R \circ \Delta_S$ are also easily shown to be equal.

From Domain of Composite Relation:
 * $\Dom {\mathcal R \circ \Delta_S} = \Dom {\Delta_S}$

But from the definition of the diagonal relation:
 * $\Dom {\Delta_S} = \Img {\Delta_S} = S$

Equality of Relations
The composite of $\Delta_S$ and $\mathcal R$ is defined as:


 * $\mathcal R \circ \Delta_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in \Delta_S \land \tuple {y, z} \in \mathcal R}$

But by definition of the diagonal relation on $S$, we have that:
 * $\tuple {x, y} \in \Delta_S \implies x = y$

Hence:
 * $\mathcal R \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y} \in \Delta_S \land \tuple {y, z} \in \mathcal R}$

But as $\forall y \in S: \tuple {y, y} \in \Delta_S$, this means:
 * $\mathcal R \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in \mathcal R}$

That is:
 * $\mathcal R \circ \Delta_S = \mathcal R$

Hence the result.

Also see

 * Diagonal Relation is Left Identity


 * Identity Mapping is Right Identity
 * Identity Mapping is Left Identity