Exercises 0.2 #5

I) S = {s1, s2, s3}, T = {t1, t2}, A = {s1}, α = {(s1, t1), (s2, t1), (s3, t2)}, α(~A) = {t1, t2}, ~α(A) = {t2}. Thus α(~A) ⊄ ~α(A).

II) Suppose α is injective and α(~A) ⊄ ~α(A). Then there is a t ∈ T such that t ∈ α(~A) and t ∉ ~α(A) ⇒ there is an s1 ∈ ~A such that α(s1) = t and t ∈ α(A) ⇒ there is an s2 ∈ A such that α(s2) = t ⇒ α(s1) = α(s2) and s1 ≠ s2. Contradiction. Thus α is injective ⇒ α(~A) ⊂ ~α(A).

III) α is surjective can’t imply α(~A) ⊄ ~α(A), otherwise a bijective map would contradict part (II). α is surjective can’t imply α(~A) ⊂ ~α(A) by the example in part (I).

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