Image of Preimage under Mapping

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

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

Corollary

 * $$B \subseteq \operatorname{Im} \left({S}\right) \implies \left({f \circ f^{-1}}\right) \left({B}\right) = B$$

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 Subset of Image 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 Subset of Image again, 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 Subset of Image 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.

Proof of Corollary
From Subset of Image we have:
 * $$B \subseteq \operatorname{Im} \left({S}\right) \implies f^{-1} \left({B}\right) \subseteq f^{-1} \left({\operatorname{Im} \left({S}\right)}\right)$$

and from Intersection with Subset is Subset we have:


 * $$f^{-1} \left({B}\right) \subseteq f^{-1} \left({\operatorname{Im} \left({S}\right)}\right) \implies f^{-1} \left({B}\right) \cap f^{-1} \left({\operatorname{Im} \left({S}\right)}\right) = f^{-1} \left({B}\right)$$

Hence the result.