Diagonal Relation is Right Identity

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

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

where $$I_S$$ is the identity mapping on $$S$$, and $$\circ$$ signifies composition of relations.

Corollary
Let $$f: S \to T$$ be a mapping.

Then:
 * $$f \circ I_S = f$$

where $$I_S$$ is the identity mapping on $$S$$, and $$\circ$$ signifies composition of mappings.

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

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

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

From Domain of Composite Relation, $$\operatorname{Dom} \left({\mathcal R \circ I_S}\right) = \operatorname{Dom} \left({I_S}\right)$$.

But from the definition of the identity mapping, $$\operatorname{Dom} \left({I_S}\right) = \operatorname{Im} \left({I_S}\right) = S$$.

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


 * $$\mathcal R \circ I_S = \left\{{\left({x, z}\right) \in S \times T: \exists y \in S: \left({x, y}\right) \in I_S \and \left({y, z}\right) \in \mathcal R}\right\}$$

But by definition of the identity mapping on $$S$$, we have that:
 * $$\left({x, y}\right) \in I_S \implies x = y$$

Hence:
 * $$\mathcal R \circ I_S = \left\{{\left({y, z}\right) \in S \times T: \exists y \in S: \left({y, y}\right) \in I_S \and \left({y, z}\right) \in \mathcal R}\right\}$$

But as $$\forall y \in S: \left({y, y}\right) \in I_S$$, this means:
 * $$\mathcal R \circ I_S = \left\{{\left({y, z}\right) \in S \times T: \left({y, z}\right) \in \mathcal R}\right\}$$

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

Hence the result.

Proof of Corollary
As a mapping is by definition also a relation, this result also holds for a mapping directly.

Also see

 * Identity Mapping is Left Identity