Identity Mapping is Surjection

Theorem

On any set $S$, the identity mapping $I_S: S \to S$ is a surjection.

Proof

The identity mapping is defined as:

$\forall y \in S: \map {I_S} y = y$

Then we have:

\(\ds \forall y \in S: \exists x \in S: \, \)

\(\ds x\)

\(=\)

\(\ds y\)

that is, $y$ itself

\(\ds \leadsto \ \ \)

\(\ds \forall y \in S: \exists x \in S: \, \)

\(\ds \map {I_S} x\)

\(=\)

\(\ds y\)

Definition of $I_S$

Hence the result.

$\blacksquare$

Also see

• Identity Mapping is Injection

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.3$