Inclusion Mapping is Surjection iff Identity

Theorem
Let $i_S: S \to T$ be the inclusion mapping.

Then:
 * $i_S: S \to T$ is surjective iff $i_S: S \to T = I_S: S \to S$

where $I_S: S \to S$ denotes the identity mapping on $S$.

Alternatively, this theorem can be worded as:
 * $i_S: S \to S = I_S: S \to S$.

It follows directly that from Surjection by Restriction of Codomain‎, the surjective restriction of $i_S: S \to T$ to $i_S: S \to \operatorname{Im} \left({i_S}\right)$ is itself the identity mapping.

Proof
It is apparent from the definitions of both the inclusion mapping and the identity mapping that:


 * $(1): \quad \operatorname{Dom} \left({i_S}\right) = S = \operatorname{Dom} \left({I_S}\right)$
 * $(2): \quad \forall s \in S: i_S \left({s}\right) = s = I_S \left({s}\right)$.

Necessary Condition
Let $i_S: S \to T = I_S: S \to S$.

From Equality of Mappings, we have that the codomain of $i_S$ equals the codomain of $I_S$.

Thus the codomain of $i_S$ equals the codomain of $I_S$ equals $S$ and thus $T = S$.

So $\forall s \in S: s = i_S \left({s}\right)$ and so $i_S$ is surjective.

Sufficient Condition
Now let $i_S: S \to T$ be a surjection.

Then $\forall s \in T: s = i_S \left({s}\right) \implies s \in S$, and therefore $T \subseteq S$.

Thus $T = S$ and so the codomain of $i_S$ equals the codomain of $I_S$ equals $S$.

Thus $i_S: S \to T = I_S: S \to S$.