Diagonal Relation is Right Identity

Theorem
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
We use the definition of mapping equality, as follows:

Equality of Domains
The ranges of $$f$$ and $$f \circ I_S$$ are equal from Range of Composite Relation:


 * $$\operatorname{Rng} \left({f \circ I_S}\right) = \operatorname{Rng} \left({f}\right) = T$$

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

From Domain of Composite Relation, $$\operatorname{Dom} \left({f \circ I_S}\right) = \operatorname{Dom} \left({I_S}\right)$$.

But from the definition of the identity mapping, $$\operatorname{Dom} \left({I_S}\right) = \operatorname{Rng} \left({I_S}\right) = S$$.

Equality of Mappings
Now we show that $$\forall x \in S: f \left({x}\right) = f \circ I_S \left({x}\right)$$.

Let $$x \in S$$.

$$ $$

Also see

 * Identity Mapping is Left Identity