Preimage of Subset under Identity Mapping

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Theorem

Let $S$ be a set.

Let $\iota_S: S \to S$ be the identity mapping on $S$.

Let $T \subseteq S$.

Then:

$\iota_S^{-1} \sqbrk T = T$

where $\iota_S^{-1} \sqbrk T$ is the preimage of $T$ under $\iota_S$.

Proof

By definition of identity mapping:

$\iota_S: S \to S: \forall x \in S: \map {\iota_S} x = x$

Let $i_S: S \to S$ be the inclusion mapping of $S$ into $S$.

By definition of inclusion mapping:

$i_S: S \to S: \forall x \in S: \map {i_S} x = x$

From Equality of Mappings:

$\iota_S = i_S$

From Preimage of Subset under Inclusion Mapping:

$\forall T\subseteq S : \iota_S^{-1} \sqbrk T = T \cap S$

From Intersection with Subset is Subset:

$\forall T\subseteq S : T \cap S = T$

Hence:

$\forall T\subseteq S : \iota_S^{-1} \sqbrk T = T$

$\blacksquare$