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

Equality of Domains

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

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

$\Box$


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 $\Dom {I_T} = T$

$\Box$


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:

$I_T \circ f = f$

$\Box$


Hence the result, by Equality of Mappings.

$\blacksquare$


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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