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Math Help - Inverse

  1. #1
    Senior Member pankaj's Avatar
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    Inverse

    Suppose X and Y are two sets and f:X\rightarrow Y is a function. For a subset A of X ,define f(A) to be the subset { {f(a):a\in A}} of Y .For a subset B of Y ,define f^{-1}(B) to be the subset { x\in X: f(x)\in B} of X.Then which of the following statements is true?

    (A) f^{-1}(f(A))=A for every A\subset X

    (B) f^{-1}(f(A))=A for every A\subset X if and only if f(X)=Y

    (C) f(f^{-1}(B))=B for every B\subset Y

    (D) f(f^{-1}(B))=B for every B\subset Y if and only if f(X)=Y

    To me it appears that every option is correct.But then I think conditions for inverse are also to be applied .

    Please explain as to why (A),(B),(C) are wrong and why (D) is correct.I am not being able to put concepts of ONE-ONE and ONTO functions in place
    Last edited by pankaj; May 3rd 2009 at 08:39 PM.
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Suppose X and Y are two sets and f:X\rightarrow Y is a function. For a subset A of X ,define f(A) to be the subset { {f(a):a\in A}} of Y .For a subset B of Y ,define f^{-1}(B) to be the subset { x\in X: f(x)\in B} of X.Then which of the following statements is true?

    (A) f^{-1}(f(A))=A for every A\subset X

    (B) f^{-1}(f(A))=A for every A\subset X if and only if f(X)=Y
    counter-example: X=\mathbb{R}, \ Y=\mathbb{R}_{+}, \ A=\{1 \}, and f: X \longrightarrow Y defined by f(x)=x^2. then f(X)=Y but f^{-1}(f(A))=\{1, -1 \} \neq A.

    \color{red} (*): as an exercise prove that (B) is true if and only if f is one-to-one!


    (C) f(f^{-1}(B))=B for every B\subset Y
    counter-example: X=Y=\mathbb{R}, \ B=\{0,-1 \}, and f: X \longrightarrow Y defined by f(x)=x^2. then f(f^{-1}(B))=\{0 \} \neq B.


    (D) f(f^{-1}(B))=B for every B\subset Y if and only if f(X)=Y
    if y \in f(f^{-1}(B)), then y=f(x), for some x \in f^{-1}(B). thus y=f(x) \in B, i.e. f(f^{-1}(B)) \subseteq B. this is true for any function f. now if f is onto and y \in B, then y=f(x), for some x \in X.

    hence f(x)=y \in B, which means x \in f^{-1}(B) and so y=f(x) \in f(f^{-1}(B)). therefore B \subseteq f(f^{-1}(B)). conversely, if f(f^{-1}(B))=B, for any B \subseteq Y, then choosing B=\{ y \}, \ y \in Y, we

    have f(f^{-1}(\{y \}))=\{y \} and thus f^{-1}(\{y \}) \neq \emptyset, i.e. f is onto.
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  3. #3
    Senior Member pankaj's Avatar
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    You make it look so simple.It is the language of mathematics that I am probably lacking in as I never studied maths formally
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  4. #4
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    In one of my very first graduate school classes, I was asked to prove, in class, some things about f^{-1}(A). I did the whole problem assuming f had an inverse and humilliated myself!
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