Subset of Codomain is Superset of Image of Preimage/Proof 2

Proof
Let $y \in B$.

Then:
 * $\exists x \in S: y = f \left({x}\right)$

Therefore by definition of preimage of subset:
 * $\exists x \in f^\gets \left({B}\right)$

It follows by definition of image of subset that:
 * $y \in f^\to \left({f^\gets \left({B}\right)}\right)$

Thus by definition of composition $f^\to$ with $f^\gets$:
 * $y \in \left({f^\to \circ f^\gets}\right) \left({B}\right)$

The result follows by definition of subset.