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Thread: Inverse

  1. #1
    Senior Member pankaj's Avatar
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    Inverse

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

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

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

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

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

    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 07:39 PM.
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Suppose $\displaystyle X$ and $\displaystyle Y$ are two sets and $\displaystyle f:X\rightarrow Y$ is a function. For a subset $\displaystyle A$ of $\displaystyle X$ ,define $\displaystyle f(A)$ to be the subset {$\displaystyle {f(a):a\in A}$} of $\displaystyle Y$ .For a subset $\displaystyle B$ of $\displaystyle Y$ ,define $\displaystyle f^{-1}(B)$ to be the subset {$\displaystyle x\in X$:$\displaystyle f(x)\in B$} of $\displaystyle X$.Then which of the following statements is true?

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

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

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


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


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

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

    have $\displaystyle f(f^{-1}(\{y \}))=\{y \}$ and thus $\displaystyle f^{-1}(\{y \}) \neq \emptyset,$ i.e. $\displaystyle f$ 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 $\displaystyle f^{-1}(A)$. I did the whole problem assuming f had an inverse and humilliated myself!
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