Preimage of Subset under Inclusion Mapping

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.