More precisely, for a function $f: X \rightarrow Y$ and sets $A \subseteq X$, $B \subseteq Y$, does $f(A) \cap f(f^{-1}(B)) = f(A) \cap B$ even though $f(f^{-1}(B)) \subseteq B$ ?
Editing to add my thoughts and some context (thanks for the feedback): after going through this list of basic properties of images and preimages as well as this question, I believe the statement to be true since $f(A) \cap f(f^{-1}(B)) = f(A \cap f^{-1}(B))$, so the intersection in question is an image which may solve the issues that come with invalidly claiming $f(A) \cap f(f^{-1}(B)) = f(A) \cap B$ since $f(f^{-1}(B)) = B$. However, I don't see how to use this fact to rigorously prove the claim.
