Image of Preimage under Mapping

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

Then:
 * $B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) = B \cap f \left({S}\right)$

Proof
As $f$ is a mapping it follows by definition that $f$ is also a relation

So we apply Preimage of Image directly:
 * $B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) \subseteq B \implies f \left({f^{-1} \left({B}\right)}\right) \subseteq B$

But from Image of Subset is Subset of Image/Corollary 3 we also have:
 * $B \subseteq T \implies f^{-1} \left({B}\right) \subseteq f^{-1} \left({T}\right)$

But from Preimage of Mapping equals Domain $f$ we have that $f^{-1} \left({T}\right) = S$ and so:
 * $B \subseteq T \implies f^{-1} \left({B}\right) \subseteq S$

Applying Image of Subset is Subset of Image/Corollary 2 we have:
 * $f^{-1} \left({B}\right) \subseteq S \implies f \left({f^{-1} \left({B}\right)}\right) \subseteq f \left({S}\right)$

From Intersection Largest it follows that:
 * $B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) \subseteq B \cap f \left({S}\right)$

Now suppose $y \in B \cap f \left({S}\right)$.

Then:
 * $f^{-1} \left({y}\right) \subseteq f^{-1} \left({B}\right)$ and $f^{-1} \left({y}\right) \subseteq f^{-1} \left({f \left({S}\right)}\right)$

and in particular:
 * $f^{-1} \left({y}\right) \subseteq f^{-1} \left({B}\right)$

Applying Image of Subset is Subset of Image/Corollary 2 again, we have:
 * $f \left({f^{-1} \left({y}\right)}\right) \subseteq f \left({f^{-1} \left({B}\right)}\right)$

But as $f$ is functional we have that:
 * $f \left({f^{-1} \left({y}\right)}\right) = y$

and so:
 * $y \in f \left({f^{-1} \left({B}\right)}\right) = \left({f \circ f^{-1}}\right) \left({B}\right)$

So we have that:
 * $B \cap f \left({S}\right) \subseteq \left({f \circ f^{-1}}\right) \left({B}\right)$

Hence the result.