Inclusion Mapping is Surjection iff Identity

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Theorem

Let $T$ be a set.

Let $S\subseteq T$ be a subset.

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


Then:

$i_S: S \to T$ is surjective if and only if $i_S: S \to T = I_S: S \to S$

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 $i_S: S \to \Img {i_S}$ is itself the identity mapping.


Proof

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

$(1): \quad \Dom {i_S} = S = \Dom {I_S}$
$(2): \quad \forall s \in S: \map {i_S} s = s = \map {I_S} s$


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 $\forall s \in S: s = \map {i_S} s$ and so $i_S$ is surjective.

$\Box$


Sufficient Condition

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


Then:

$\forall s \in T: s = \map {i_S} s \implies s \in S$

and therefore:

$T \subseteq S$

Thus:

$T = S$

and so the codomain of $i_S$ equals the codomain of $I_S$ which equals $S$.


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

$\blacksquare$


Sources

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