Identity Mapping is Right Identity/Proof 1

Theorem
Let $S$ and $T$ be sets.

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.

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, $\operatorname{Dom} \left({f \circ I_S}\right) = \operatorname{Dom} \left({I_S}\right)$.

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

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


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

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

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

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

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

Hence the result, by Equality of Mappings.