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 can use the definition of mapping equality.

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


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

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

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

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

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

Let $$x \in S$$.

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