Identity Mapping is Surjection

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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:

\(\displaystyle \forall y \in S: \exists x \in S: x\) \(=\) \(\displaystyle y\) that is, $y$ itself
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall y \in S: \exists x \in S: \map {I_S} x\) \(=\) \(\displaystyle y\) Definition of $I_S$

Hence the result.

$\blacksquare$


Also see


Sources