# How to prove $f^{-1}(f(A))=A \quad \Longrightarrow f(A\cap B)=f(A)\cap B$

$$X,Y$$ are Quantites and $$f:X\rightarrow Y$$ a function that is injective.

i have already proven that $$f^{-1}(f(A))=A$$ when $$f$$ is injective.

How to prove $$f^{-1}(f(A))=A \quad \Longrightarrow f(A\cap B)=f(A)\cap B$$?

• Perhaps you mean $f(A \cap B) = f(A) \cap f(B)$. – Robert Israel Apr 24 at 19:42
• Even with @RobertIsrael 's correction, I'm not convinced this is a true statement. Also, $X,Y$ are sets, not quantities. – Don Thousand Apr 24 at 19:44
• Hello Don Thousand, i have to show the equivalence of five statements. And I thought that i'd proof the equivalence like this: a => b => c => d => e => a – Analysis Apr 24 at 19:45
• Maybe this is worth reading: $f(A\cap B)=f(A)\cap f(B)$ $\iff$ $f$ is injective. – Minus One-Twelfth Apr 24 at 19:48
• @MinusOne-Twelfth This is probably what OP meant. I would close as a duplicate of that problem – Don Thousand Apr 24 at 19:49

Yes, $$f^{-1}(f(A))=A$$
if for all A, $$f^{-1}(f(A))=A,$$