Identity Mapping is Right Identity/Proof 1

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

Equality of Domains
From Domain of Composite Relation:
 * $\Dom {f \circ I_S} = \Dom {I_S}$

But from the definition of the identity mapping:
 * $\Dom {I_S} = \Img {I_S} = S$

Equality of Mappings
The composite of $I_S$ and $f$ is defined as:


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

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

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

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

That is:
 * $f \circ I_S = f$

Hence the result, by Equality of Mappings.