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

By definition of preimage:

$i_H^{-1} \sqbrk T = \set {h \in H: \exists t \in T: \map {i_H} h = t}$

Let $T \cap H = \O$.

Then:

$\nexists h \in H: \exists t \in T: \map {i_H} h = t$

That is:

$i_H^{-1} \sqbrk T = \O$

That is:

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

$\Box$

Let $T \cap H \ne \O$.

From Intersection is Subset:

$H \ne \O$

and:

$T \ne \O$

From Subset of Empty Set:

$S \ne \O$

Suppose $i_H^{-1} \sqbrk T = \O$.

Then from Empty Set is Subset of All Sets:

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

Otherwise $i_H^{-1} \sqbrk T \ne \O$.

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

By the definition of inclusion mapping:

$\map {i_H} x = x$

By definition of preimage of $T$ under $i_H$:

$\map {i_H} x \in T$

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 by definition of inclusion mapping again:

$\map {i_H} x \in T$

and so by definition of preimage of $T$:

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

Therefore:

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

Hence, by definition of subset:

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

$\Box$

We have shown that:

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

and:

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

Hence by definition of set equality:

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

$\blacksquare$