Proposition 0.3.15. Consider f : A - B. Let C, D be subsets of B. Then f-' ( CUD ) = f-'(C) uf-' (D), f(CnD) = f-' (cinf-' ( D), f-(0 ) = (f-'(0))". 0.3. BASIC SET THEORY 15 Read the last line of the proposition as f-1(B \\ C) =A\\f-1(C). Proof. Let us start with the union. Suppose x e f (CUD), meaning that x is taken to C or D. Thus f-'(CUD) cf-(C)Uf-1(D). Conversely if xef-'(C), then x e f-'(CUD). Similarly for x e f (D). Hence f '(CUD) > f (C)Uf (D), and we have equality. The rest of the proof is left as an exercise. 0 The proposition does not hold for direct images. We do have the following weaker result. Activate