Identity Mapping is Right Identity

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.

Proof 1
We use the definition of mapping equality, as follows:

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.

Proof 2
By definition, a mapping is also a relation.

Also by definition, the identity mapping is the same as the diagonal relation.

Thus Diagonal Relation is Right Identity can be applied directly.

Also see

 * Identity Mapping is Left Identity


 * Diagonal Relation is Right Identity
 * Diagonal Relation is Left Identity