a) Show that if f:A->B is injective and E is a subset of A, then f -1 (f(E))=E.
ID: 2940692 • Letter: A
Question
a) Show that if f:A->B is injective and E is a subset of A, then f-1(f(E))=E. Give an example to show that equality need not hold if f is not injective. b) Show taht if f:A-B is surjective and H is a subset of B, then f(f-1(H))=H. Give an example to show that equality need not hold if f is not surjective. a) Show that if f:A->B is injective and E is a subset of A, then f-1(f(E))=E. Give an example to show that equality need not hold if f is not injective. b) Show taht if f:A-B is surjective and H is a subset of B, then f(f-1(H))=H. Give an example to show that equality need not hold if f is not surjective.Explanation / Answer
Question Details a) Show that if f:A->B is injective[ONE TO ONE] and E is a subset of A, then f-1(f(E))=E. Give an example to show that equality need not hold if f is not injective. GIVEN F[A]=B F IS INJECTIVE E IS A SUBSET OF A TST F^(-1)[F(E)]=E LET P BE ANY ELEMENT OF E E IS SUB SET OF A HENCE P IS AN ELEMENT OF A LET F[P]=Q WHERE Q IS AN ELEMENT OF B, SINCE F[A]=B AND F IS INJECTIVE, THERE SHALL BE UNIQUE ELEMENT Q IN B SUCH THAT F[P]=Q ,WHERE P IS ALSO AN UNIQUE ELEMENT IN A. THAT IS F INVERSE [Q]=P, WHERE P IS ANY ELEMENT IN E SO THIS LOGIC HOLDS GOOD FOR ALL ELEMENTS IN E SO F INVERSE[F(E)]=E THIS WILL NOT BE SO IF F IS NOT INJECTIVE TAKE A=[1,-1,2,-2,.....]..............E=[1] B=[1,4,....] F[A]=A*A=B THEN FOR INVERSE OF 1 WE GET BOTH +1 AND -1,WHERE AS E CONSISTS OF [1] ONLY.. b) Show taht if f:A-B is surjective[ON TO ] and H is a subset of B, then f(f-1(H))=H. Give an example to show that equality need not hold if f is not surjective. GIVEN F[A]=B F IS SURJECTIVE H IS A SUBSET OF B TST F^(-1)[F(H)]=H LET P BE ANY ELEMENT OF H H IS SUB SET OF B HENCE P IS AN ELEMENT OF B LET F[Q]=P WHERE P IS AN ELEMENT OF A . SINCE F[A]=B AND F IS SURJECTIVE, THERE SHALL BE UNIQUE ELEMENT Q IN A SUCH THAT F[Q]=P ,WHERE P IS ALSO AN UNIQUE ELEMENT IN B. THAT IS F INVERSE [P]=Q, WHERE P IS ANY ELEMENT IN H SO THIS LOGIC HOLDS GOOD FOR ALL ELEMENTS IN H SO F INVERSE[F(H)]=H THIS WILL NOT BE SO IF F IS NOT SURJECTIVE TAKE A=[1,-1,2,-2....] B=[1,-4,4......]..............H=[4] F[A]=A*A=B THEN FOR INVERSE OF -4 WE DO NOT GET ANY CORRESPONDING ELEMENT IN A.
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