Identity Mapping is Left Identity/Proof 1

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


 * $\Dom {I_T \circ f} = \Dom f$

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

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

But from the definition of the identity mapping, the codomain of $I_T$ is $\Dom {I_T} = T$

Equality of Relations
The composite of $f$ and $I_T$ is defined as:


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

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

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

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

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

Hence the result, by Equality of Mappings.