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Thread: Composition/Image Proof

  1. #1
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    Composition/Image Proof

    If f:A→B is 1-1 and im f=B prove that (f^-1 ○f)(a)=a for all a element of A and (f^-1 ○f)(b)=b for each b an element of B.


    Here is what I've got out of this.
    f is 1-1, so for each b, there is one and only a. So $\displaystyle f^-1(b)=a$
    Im f=b. Thus, f is onto. Therefore, the range is equal to the set B. Would this imply f(a)=b and f(b)=b?

    I'm just not sure where to go with this problem.
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  2. #2
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    Suppose $\displaystyle f(x) = y$.
    Then $\displaystyle f^{-1}(y) = x$.

    So$\displaystyle f^{-1}(f(x)) = f^{-1}(y) = x$.

    Similarly,
    $\displaystyle f^{-1}(y) = x$ gives us
    $\displaystyle f(x) = y$.

    So $\displaystyle f(f^{-1}(y)) = f(x) = y$.
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    Suppose $\displaystyle f(x) = y$.
    Then $\displaystyle f^{-1}(y) = x$.

    So$\displaystyle f^{-1}(f(x)) = f^{-1}(y) = x$.

    Similarly,
    $\displaystyle f^{-1}(y) = x$ gives us
    $\displaystyle f(x) = y$.

    So $\displaystyle f(f^{-1}(y)) = f(x) = y$.
    How does the im F=B work into the proof? I don't get that part.
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  4. #4
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    Quote Originally Posted by kathrynmath View Post
    How does the im F=B work into the proof? I don't get that part.
    One reason I did not respond before now is that I consider this question to have notational problems. Those problems are reflected in the icemanfanís answer, he assumed the ordinary meaning of the symbols.

    However is these set theory proofs, $\displaystyle f^{ - 1} (b) = \left\{ {x \in A:f(x) = b} \right\}$. That is the inverse image is a set.
    The fact that $\displaystyle f$ is a bijection simply means that $\displaystyle \left( {\forall b \in B} \right)\left[ {f^{ - 1} (b) \ne \emptyset } \right]$.
    By injectivity $\displaystyle \left\{ {p,q} \right\} \subseteq f^{ - 1} (b) \Rightarrow \quad f(p) = f(q) = b \Rightarrow \quad p = q$.
    Which means that each $\displaystyle f^{-1}(b)$ is a singleton set.
    So $\displaystyle \left( {\forall a \in A} \right)\left[ {f(a) \in B} \right] \Rightarrow \quad \left\{ a \right\} = f^{ - 1} \left( {f(a)} \right)$.

    Now I understand that this notation is not what the question implies.
    However, I hope it may give you some guidance.
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  5. #5
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    Quote Originally Posted by kathrynmath View Post
    How does the im F=B work into the proof? I don't get that part.
    That just means that all of the elements of B are values that can be output by F. Hence, F is a bijection, and all elements of B have inverses of F in A.
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  6. #6
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    Thanks, I think I got the right idea to finish the problem now.
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