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It follows by definitic ecause F is onto, there exists an x in X, such that F(x) = y image of a set that F(x) E F-(B) . So, since y EB and by definition of an inverse image, x € F-"(B) Therefore, by substitution, y e F(F-"(B)) onclusion: Since both subset relations have been proved, we conclude that F(F¯1 (B)) = B.

It follows by definitic ecause F is onto, there exists an x in X, such that F(x) = y image of a set that F(x) E F-(B) . So, since y EB and by definition of an inverse image, x € F-"(B) Therefore, by substitution, y e F(F-"(B)) onclusion: Since both subset relations have been proved, we conclude that F(F¯1 (B)) = B.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.2: Mappings
Problem 27E: 27. Let , where and are nonempty. Prove that has the property that for every subset of if and...
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Let X and Y be any sets, and let F be any onto function from X to Y. Prove that for every subset B C Y, F(F¯1 (B)) = B. Proof: To prove that F(F (B)) B, we must show that F(F` (B)) C B and that B C F(F¯1 (B)). Proof that F(F¯1 (B)) C B: Suppose y E F(F1(B)). We must show that y e B [we By definition of image of a set, there exists an element x in F-(B) such that F(x) = y . In addition, because x is in F-(B) , by definition of inverse image, F(x) E B Therefore, by substitution, y € B Proof that B C F(F¯1 (B)): Suppose y E B. We must show that y is in F(F-"(B)) It follows by definition Because F is onto, there exists an x in X, such that F(x) = y of image of a set that F(x) E F-(B) So, since y E B and by definition of an inverse image, x E F-(B) Therefore, by substitution, y E F(F-(B)) Conclusion: Since both subset relations have been proved, we conclude that F(F (B)) = B.
Transcribed Image Text:Let X and Y be any sets, and let F be any onto function from X to Y. Prove that for every subset B C Y, F(F¯1 (B)) = B. Proof: To prove that F(F (B)) B, we must show that F(F` (B)) C B and that B C F(F¯1 (B)). Proof that F(F¯1 (B)) C B: Suppose y E F(F1(B)). We must show that y e B [we By definition of image of a set, there exists an element x in F-(B) such that F(x) = y . In addition, because x is in F-(B) , by definition of inverse image, F(x) E B Therefore, by substitution, y € B Proof that B C F(F¯1 (B)): Suppose y E B. We must show that y is in F(F-"(B)) It follows by definition Because F is onto, there exists an x in X, such that F(x) = y of image of a set that F(x) E F-(B) So, since y E B and by definition of an inverse image, x E F-(B) Therefore, by substitution, y E F(F-(B)) Conclusion: Since both subset relations have been proved, we conclude that F(F (B)) = B.
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