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If A and B are sets and f : A + B, then for any subset S of A we define f(S) = {b B

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If A and B are sets and f : A + B, then for any subset S of A we define f(S) = {b B : b = f(a) for some a S}. Similarly, for any subset T of B we define the pre-image of T as (T) = {a A: f(a) T} Note that f-'(T) is well defined even if f does not have an inverse ! For each of the following state whether it is True or False. If True then give a proof. If False then give a counterexample: (a) f(SUS) = f(Si) u f(S2) (b) f(Sin S2) = f(S) n (S2) (c) f-'(Tu Tx) = f'(Ti) u f-(T2) (d) f-'(Tin Tx) = f'(Tu) nf-'(T2)

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