Let A be a subset of a set X, let B be a subset of a set Y and let f:X->Y (so f
ID: 3077653 • Letter: L
Question
Let A be a subset of a set X, let B be a subset of a set Yand let f:X->Y (so f is a function from X to Y, dom f = X,
and the range of f is a subset of Y).
The text proves that A is a subset of f^{-1}(f(A)), with equality
if f is injective.
(1)Prove that the "if" can be replaced by "if and only if" if the
equality holds for all subsets A of X.
The text also proves that f(f^{-1}(B)) is a subset of B, with
equality if f is surjective.
(2)Prove that the "if" can be replaced by "if and only if" if the
equality holds for all subsets B of Y.
Hand write clearly please :)
Explanation / Answer
1) Let the equality hold for all subsets A of X. Assume f in not injective. We hope to find a contradiction.
By our assumption there exists two distinct elements x_1 and x_2 in X such that f(x_1)=f(x_2)=y ,say.
Take A={x_1}.
Then f(A)=y, and f^{-1}(f(A))=f^{-1}(y)={x_1,x_2}.
But its clear that {x_1,x_2} not equals {x_1}.
That is A does not equal f^{-1}(f(A)), which is a contradiction and we are done.
2) Let the equality hold for all subsets B of Y and assume that f isn't surjective.
We hope to produce a contradiction.
By assumption there exists y in Y such that f(x) does not map to y for any damn x.
Take B={y}. Then f^{-1}(B)=. So f(f^{-1}(B))=.
Clealy does not eqaul {y}. Thus f(f^{-1}(B)) does not equal B, which is a contradiction.
Hence we are done.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.