Identity Mapping is a Surjection

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Theorem

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

The identity mapping is defined as $\forall y \in S: I_S \left({y}\right) = y$. Then we have:

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \forall y \in S: \exists x \in S: x$	$=$	$\displaystyle y$	$\displaystyle $	$\displaystyle $	$\displaystyle $	(i.e. $y$ itself)
	$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \forall y \in S: \exists x \in S: I_S \left({x}\right)$	$=$	$\displaystyle y$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by definition

Hence the result.

$\blacksquare$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \forall y \in S: \exists x \in S: x\)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(i.e. $y$ itself)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \forall y \in S: \exists x \in S: I_S \left({x}\right)\)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition