Identity Mapping is Right Identity

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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

Equality of Codomains

The codomains of $f$ and $f \circ I_S$ are both equal to $T$ from Codomain of Composite Relation.

$\Box$


Equality of Domains

From Domain of Composite Relation:

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

But from the definition of the identity mapping:

$\Dom {I_S} = \Img {I_S} = S$

$\Box$


Equality of Mappings

The composite of $I_S$ and $f$ is defined as:

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

But by definition of the identity mapping on $S$, we have that:

$\tuple {x, y} \in I_S \implies x = y$

Hence:

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


But as $\forall y \in S: \tuple {y, y} \in I_S$, this means:

$f \circ I_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in f}$

That is:

$f \circ I_S = f$

$\Box$


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 Right Identity can be applied directly.

$\blacksquare$


Also see


Sources

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