Identity Mapping is Left Identity

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

From Range of Composite Relation, $$\operatorname{Rng} \left({I_T \circ f}\right) = \operatorname{Rng} \left({I_T}\right) = T$$.

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

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

Let $$x \in S$$.

$$ $$

Also see

 * Identity Mapping is Right Identity