Preimage of Image of Subset under Injection equals Subset

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

Then:
 * $\forall A \in \mathcal P \left({S}\right): A = \left({f^\gets \circ f^\to}\right) \left({A}\right)$

where:


 * $f^\to$ denotes the mapping induced on the power set $\mathcal P \left({S}\right)$ of $S$ by $f$
 * $f^\gets$ denotes the mapping induced on the power set $\mathcal P \left({T}\right)$ of $T$ by the inverse $f^{-1}$
 * $f^\gets \circ f^\to$ denotes composition of $f^\gets$ and $f^\to$.

Proof
Let $f$ be an injection.

From Subset of Domain is Subset of Preimage of Image, we have that:
 * $\forall A \in \mathcal P \left({S}\right): A \subseteq \left({f^\gets \circ f^\to}\right) \left({A}\right)$

by dint of $f$ being a relation.

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

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

Let $x \in A$.

We have:

Thus we see that:
 * $\left({f^\gets \circ f^\to}\right) \left({A}\right) \subseteq A$

and hence the result:
 * $\forall A \in \mathcal P \left({S}\right): A = \left({f^\gets \circ f^\to}\right) \left({A}\right)$

Also see

 * Subset equals Preimage of Image implies Injection
 * Subset equals Preimage of Image iff Mapping is Injection