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

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

From Equality of Mappings, we have that $$\operatorname{Rng} \left({i_S}\right) = \operatorname{Rng} \left({I_S}\right)$$ are equal.

Thus $$\operatorname{Rng} \left({i_S}\right) = \operatorname{Rng} \left({I_S}\right) = S$$ 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 $$\operatorname{Rng} \left({i_S}\right) = S = \operatorname{Rng} \left({I_S}\right)$$.

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