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  1. #1
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    surjective proof

    I have no idea how to prove the following, any help on how to resolve this would be appreciated:

    Let f be a function from X onto Y and let B \subset Y. Show that f(f^{-1}(B)) = B.
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    Quote Originally Posted by lllll View Post
    I have no idea how to prove the following, any help on how to resolve this would be appreciated:

    Let f be a function from X onto Y and let B \subset Y. Show that f(f^{-1}(B)) = B.
    You need to show two things.
    1)If x\in f(f^{-1}(B)) then x\in B.
    2)If x\in B then x\in f(f^{-1}(B)).
    Together with these steps it shall prove f(f^{-1}(B)) = B.

    I start you off.
    If x\in f(f^{-1}(B)) then x = f(y) for some y\in f^{-1}(B).
    But y \in f^{-1}(B) means f(y) \in B, by definition, thus x\in B.

    Now you go the other way.
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  3. #3
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    Quote Originally Posted by lllll View Post
    I have no idea how to prove the following, any help on how to resolve this would be appreciated:

    Let f be a function from X onto Y and let B \subset Y. Show that f(f^{-1}(B)) = B.
    By definition f^{-1}(B) ={ x:xεX & f(x)εB }.....................................1

    So we see that  f^{-1}(B) IS A subset of X. Now since for any subset ,lets say A, of X we have by definition f(A) ={ f(x) : xεA } then  f( f^{-1}(B)) ={ f(x): xε  f^{-1}(B) }................................................. ...........2



    So we see that  f( f^{-1}(B)) is a set of the images of x f(x) , such that xε  f^{-1}(B).


    The question now is are these f(x)'s inside B??

    Lets see:

    Let .........................f(x)ε  f( f^{-1}(B)) then from 2 we infer f(x)=f(x) & xε  f^{-1}(B) .

    But from 1 if xε  f^{-1}(B) we conclude xεX & f(x)εB.

    Hence f(x) is inside B.AND  f( f^{-1}(B)) is a subset of B

    Now for the converse let f(x)εB. Since f is onto there exists an xεX such that f(x)εB AND so by 1 xε f^{-1}(B)........................e.t.c,..........e.t.c
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    You need to show two things.
    1)If x\in f(f^{-1}(B)) then x\in B.
    2)If x\in B then x\in f(f^{-1}(B)).
    Together with these steps it shall prove f(f^{-1}(B)) = B.

    I start you off.
    If x\in f(f^{-1}(B)) then x = f(y) for some y\in f^{-1}(B).
    But y \in f^{-1}(B) means f(y) \in B, by definition, thus x\in B.

    Now you go the other way.
    For the other way around, does this work?

    Say x \in B \subset Y
    Then there exists an y \in f^{-1}(B) \subset X such that f(y)=x, by definition of a surjective function
    But then: f(y) \in f( f^{-1}(B)) and since x=f(y), we have x \in f( f^{-1}(B)) as desired.
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  5. #5
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    Quote Originally Posted by Last_Singularity View Post
    For the other way around, does this work?

    Say x \in B \subset Y
    Then there exists an y \in f^{-1}(B) \subset X such that f(y)=x, by definition of a surjective function
    But then: f(y) \in f( f^{-1}(B)) and since x=f(y), we have x \in f( f^{-1}(B)) as desired.
    Yes it works.
    It indeed has to be a surjective function;
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