Subset equals Image of Preimage iff Mapping is Surjection

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

iff $$f$$ is a surjection.

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

In particular, it holds for $$T$$ itself.

Hence:

$$ $$ $$ $$

So:
 * $$T \subseteq \operatorname{Im} \left({f}\right) \subseteq T$$

and so by Equality of Sets:
 * $$\operatorname{Im} \left({f}\right) = T$$

So, by definition, $$f$$ is a surjection.

Necessary Condition
Let $$f$$ be a surjection.

Let $$B \subseteq T$$.

Let $$b \in B$$.

Then:

$$ $$ $$ $$

From Preimage of Image, we already have that:
 * $$f_{g} \left({f_{g^{-1}} \left({B}\right)}\right) \subseteq B$$

So:
 * $$B \subseteq f_{g} \left({f_{g^{-1}} \left({B}\right)}\right) \subseteq B$$

and from Equality of Sets:
 * $$B f_{g} \left({f_{g^{-1}} \left({B}\right)}\right)$$