# Inclusion Mapping is Surjection iff Identity

## 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

- 1960: Paul R. Halmos:
*Naive Set Theory*... (previous) ... (next): $\S 8$: Functions