Prove or disprove $f^{-1}(A - B) = f^{-1}(A)-f^{-1}(B)$

Question

I'm getting that we can neither prove nor disprove it, but it seems there's a definitive answer for this. How to prove or disprove this?

Answer 1

This is true since preimage commutes with set difference.

Theorem: Let $A$ and $B$ be sets, and consider a function $f: A \rightarrow B .$ Let $B_{1} \subseteq B$ and $B_{2} \subseteq B$ be any subsets. Then we have $$ f^{-1}\left(B_{1} \backslash B_{2}\right)=f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right) $$ Proof: For any given $x \in A$, we have $$ \begin{aligned} x \in f^{-1}\left(B_{1} \backslash B_{2}\right) & \Longleftrightarrow f(x) \in B_{1} \backslash B_{2} \\ & \Longleftrightarrow f(x) \in B_{1} \wedge f(x) \notin B_{2} \\ & \Longleftrightarrow x \in f^{-1}\left(B_{1}\right) \wedge x \notin f^{-1}\left(B_{2}\right) \\ & \Longleftrightarrow x \in f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right) \end{aligned} $$ and therefore, since they comprise the same elements of $A$, $$ f^{-1}\left(B_{1} \backslash B_{2}\right)=f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right) $$