Preimage of Image of Subset under Injection equals Subset

Theorem
Let $g: S \to T$ be an injection.

Let $f_g: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$ be the mapping induced by $g$.

Similarly, Let $f_{g^{-1}}: \mathcal P \left({T}\right) \to \mathcal P \left({S}\right)$ be the mapping induced by the inverse $g^{-1}$.

Then:
 * $\forall A \in \mathcal P \left({S}\right): A = \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right)$

Proof
Let $f$ be an injection.

From Subset of Domain is Subset of Preimage of Image, we have that:
 * $\forall A \in \mathcal P \left({S}\right): A \subseteq \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right)$

by dint of $f$ being a relation.

So what we need to do is show that:
 * $\forall A \in \mathcal P \left({S}\right): \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right) \subseteq A$

Take any $A \in \mathcal P \left({S}\right)$.

Let $x \in A$.

We have:

Thus we see that:
 * $\left({f_{g^{-1}} \circ f_g}\right) \left({A}\right) \subseteq A$

and hence the result:
 * $\forall A \in \mathcal P \left({S}\right): A = \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right)$