Subset equals Preimage of Image iff Mapping is Injection

Theorem
Let $$g: S \to T$$ be a mapping.

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

iff $$f$$ is an injection.

Sufficient Condition
Let $$f$$ be such that:
 * $$\forall A \in P \left({S}\right): A = \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right)$$

In particular, it holds for all subsets of $$A$$ which are singletons.

Now, consider any $$x, y \in A$$.

We have:

$$ $$ $$ $$

So $$f$$ is an injection.

Necessary Condition
Let $$f$$ be an injection.

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

by dint of $$f$$ being a relation.

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

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

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