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Und f(b) = z mit y 6= z Dann gehört zwar das Element y zu f(A)∩f(B) = {y}, es gehört aber nicht zu f(A ∩ B) = f({b}) = {z} ( ) Nach der Definition von f −1ist f (f(A)) = {x ∈ X f(x) ∈ f(A)} Gehört nun ein Element x zu der Menge A, so gehört auch f(x) zu dem Bild B = f(A) von A Dann muss aber das Element x zu dem Urbild f−1(B) = f−1(f(A)) gehören Die umgekehrte Inklusio. 2 / 2 3 1 3 = ^ X _ Y X ` Z a Q V a b Q Z ` ^ X W c R d R e X f _ X g h \ a V i Q W j > = ?. J f b g k ?.

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