Identity Mapping is Left Identity/Proof 1

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

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.

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


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

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 $\operatorname{Dom} \left({I_T}\right) = T$

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


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

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

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

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

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

Hence the result, by Equality of Mappings.