Preimage of Subset under Inclusion Mapping

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Theorem

Let $S$ be a set.

Let $H \subseteq S$ be a subset of $S$.

Let $i_H: H \to S$ be the inclusion mapping on $H$.


Let $T \subseteq S$.

Then:

$i_H^{-1} \sqbrk T = T \cap H$

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


Proof

Let $x \in i_H^{-1} \sqbrk T$.

By the definition of inclusion mapping:

$\map {i_H} x = x$

By definition of preimage of $T$:

$\map {i_H} x \in T$

Thus $x \in T$.

By definition of inclusion mapping:

$x \in \Dom {i_H} = H$

So $x \in T$ and $x \in H$ and so by definition of set intersection:

$x \in T \cap H$

Thus by definition of subset:

$i_H^{-1} \sqbrk T \subseteq T \cap H$


Now let $x \in T \cap H$.

As $H = \Dom {i_H}$ it follows that:

$x \in \Dom {i_H}$

As $x \in T \cap H$ it follows by definition of set intersection that $x \in T$.

Thus:

$\map {i_H} x \in T$

and so by definition of preimage of $T$:

$x \in i_H^{-1} \sqbrk T$

Thus by definition of subset:

$T \cap H \subseteq i_H^{-1} \sqbrk T$


The result follows by definition of set equality.

$\blacksquare$