I) S = {s_{1}, s_{2}, s_{3}}, T = {t_{1}, t_{2}}, A = {s_{1}}, α = {(s_{1}, t_{1}), (s_{2}, t_{1}), (s_{3}, t_{2})}, α(~A) = {t_{1}, t_{2}}, ~α(A) = {t_{2}}. 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 s_{1} ∈ ~A such that α(s_{1}) = t and t ∈ α(A) ⇒ there is an s_{2} ∈ A such that α(s_{2}) = t ⇒ α(s_{1}) = α(s_{2}) and s_{1} ≠ s_{2}. 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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