Identity Mapping is Left Identity

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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.


Proof 1

We use the definition of mapping equality, as follows:


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.

$\blacksquare$


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 Left Identity can be applied directly.

$\blacksquare$


Also see


Sources

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