Subset equals Preimage of Image implies Injection/Proof 2

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 $g^{-1}$.

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

Then $g$ is an injection.

Proof
Suppose that $g$ is not an injection.

Then two elements of $S$ map to the same one element of $T$.

That is:
 * $\exists a_1, a_2 \in S, b \in T: g \left({a_1}\right) = g \left({a_2}\right) = b$

Let $A = \left\{{a_1}\right\}$.

Then:

So by the Rule of Transposition:
 * $\forall A \in \mathcal P \left({S}\right): A = \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right)$

implies that $g$ is an injection.