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Conversely, assume that f(f−1) = C for all C ⊆ Y but that f is not ontoThen there exists y∈ Y such that for all x∈ X, we have y6= f(x) Let C= {y} Then f−1 = ∅ and f(f−1) = ∅ 6= C since y∈ C ContradictionHence, f is onto Thus, f is onto if and only if f(f−1) = C for all C⊆ Y 8(a) Claim f(A∪B) = f(A) ∪f(B)T v ÁF ʎВc @ l m t C O f B X N ÁF ЃN u W j A ǁF S r ` f B X N t F X e B o s ψ F N7 18 ( y) @1300 @ ` @7 19 ( )C G t X R ^ { g i q s f @ P O X j U I X N ₩ ȉ F ̃{ g ɂd r c V { t C G t X R ^ { g ͐ ~ ɂȂ ܂ B ݌Ɍ ł ̂ł B

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Mean Value Theorem Wikipedia

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Injective Function Wikipedia

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