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 Image of Preimage is Subset 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 is Largest Subset 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.