Inclusion Mapping is Surjection iff Identity

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

Then $$i: S \to T$$ is surjective iff $$i: S \to T = I_S: S \to S$$.

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

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

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


 * 1) $$\mathrm{Dom} \left({i}\right) = S = \mathrm{Dom} \left({I_S}\right)$$;
 * 2) $$\forall s \in S: i \left({s}\right) = s = I_S \left({s}\right)$$.


 * Let $$i: S \to T = I_S: S \to S$$.

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

Thus $$\mathrm{Rng} \left({i}\right) = \mathrm{Rng} \left({I_S}\right) = S$$ and thus $$T = S$$.

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


 * Now let $$i: S \to T$$ be a surjection.

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

Thus $$T = S$$ and so $$\mathrm{Rng} \left({i}\right) = S = \mathrm{Rng} \left({I_S}\right)$$.

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