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


Sources

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