Identity Mapping is Left Identity

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

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

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

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

Then:
 * $I_T \circ f = f$

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

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

Equality of Domains
The domains of $\mathcal R$ and $I_T \circ \mathcal R$ are equal from Domain of Composite Relation:


 * $\operatorname{Dom} \left({I_T \circ \mathcal R}\right) = \operatorname{Dom} \left({\mathcal R}\right)$

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

From Codomain of Composite Relation, the codomains of $I_T \circ \mathcal R$ and $I_T$ are both equal to $T$.

But from the definition of the identity mapping, the codomain of $I_T$ is $\operatorname{Dom} \left({I_T}\right) = T$

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


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

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

Hence:
 * $I_T \circ \mathcal R = \left\{{\left({x, y}\right) \in S \times T: \exists y \in T: \left({x, y}\right) \in \mathcal R \land \left({y, y}\right) \in I_T}\right\}$

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

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

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 Right Identity