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Math Help - Function and its inverse

  1. #1
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    Function and its inverse

    Dear Friends ,
    The following points can be proved easily :-
    a) A function has a left inverse if and only if it is one-one.
    b) A function has a right inverse if and only if it is onto.

    A function is said to have an inverse if and only if its left and right inverses exist and are equal to each other.
    We also know that a function has an inverse only if it is bijective. However the if part is never said. That is , it is never said that a function has an inverse if and only if it is bijective. We remain silent about the if part.
    The question is that is it possible for a bijective function to have left and right inverses that are unequal ? Putting it into another way , the question is that is it possible for a function to be bijective and still not have an inverse ?

    Please reply and point out any error in the argument if any .
    Thanks
    UJJWAL
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  2. #2
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    I'll phrase your question in a different way - given f: A \to B is bijective, can you construct a function g:B \to A such that g is bijective and is the inverse of f?
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  3. #3
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    @Defunkt ,

    To be honest i did not understand your phrasing. I hope you would make it clearer. I would like to know how it can be phrased as such.
    By the way I seem to have figured out the solution. I am putting it down.
    Let me put down some basic results that can be proved easily :-
    a) A function has got a left inverse iff it is one-one. Moreover, that left inverse is unique over the co-domain of the original function.
    b) A function has got a right inverse iff it is surjective. Moreover that right inverse is unique over the co-domain of the original function.
    Now suppose that we have a function f:X->Y. Suppose that f is bijective. So let g:Y->X be its left inverse and h:Y->X be its right inverse. Also suppsoe that there exists one yi in the set Y such that g(yi) is not equal to h(yi).
    Now suppose g(yi)=xe1. It is easy to see that f(xe1)=yi otherwise g will not be the left inverse of f.
    Also suppose that h(yi)=xe2.
    Now foh (yi) = f(h(yi))=f(xe2) = yi (From the assumption that h is the right inverse of f ).
    But f(xe1)= yi.
    So we have f(xe1) = f(xe2) = yi and since f is bijective so we must have xe1 = xe2 . Hence our assumption is invalid and hence the left and the right inverse must be equal.
    From here we conclude that if a function is bijective it will necessarily have a left inverse as well as a right inverse and they both shall necesarily be equal to each other, thus making the function possess a unique inverse.

    I have just sketched this proof. Kindly go through it and feel free to pass any comments or suggestions that you may have.

    Yours

    UJJWAL
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  4. #4
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    You asked
    is it possible for a function to be bijective and still not have an inverse ?
    I give you a bijective function f: A \to B

    That means that if I take any b \in B, there exists a unique  a \in A such that f^{-1}(b) = a.
    That means that the inverse f^{-1} of f is well defined and exists for any b \in B, which is pretty much what you need.

    I read through your proposed proof, the notation is a little unclear but I think it's alright.
    Last edited by Defunkt; August 29th 2010 at 02:38 PM. Reason: tex
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  5. #5
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    Thank you for reading my proof. I understand it would be a bit unclear because i don't know about Latex so could not put it in a nice mathematical form. I am learning it and hopefully things would change.

    I indeed needed the answer to whether i can construct the inverse function or not. The thing is that i needed the answer very rigorously. You know many people mistake the concept of an inverse function. The inverse of a function is defined in terms of the left inverse and the right inverse.
    You know in the above post . I can define a number of f^{-1} all of which will be from B to A and will be well defined. But it is not necessary that all of them will be the inverse. I needed the answer that the left and the right inverse are necessarily equal when a function is bijective. I hope now I have got that. So my version and your version of the problem are different.
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  6. #6
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    Quote Originally Posted by UJJWAL View Post
    Thank you for reading my proof. I understand it would be a bit unclear because i don't know about Latex so could not put it in a nice mathematical form. I am learning it and hopefully things would change.

    I indeed needed the answer to whether i can construct the inverse function or not. The thing is that i needed the answer very rigorously. You know many people mistake the concept of an inverse function. The inverse of a function is defined in terms of the left inverse and the right inverse.
    The definition I was using is an equivalent one - it does not have to be defined with left and right inverses.

    You know in the above post . I can define a number of f^{-1} all of which will be from B to A and will be well defined. But it is not necessary that all of them will be the inverse. I needed the answer that the left and the right inverse are necessarily equal when a function is bijective. I hope now I have got that. So my version and your version of the problem are different.
    Yes, but your function is f^{-1}: B \to \mathbb{P}(A) which assigns a set to each element b \in B.
    The difference is that since f is injective, you can be assured that each b \in B will get matched with at most one a \in A.
    And since f is surjective, you can be assured that each b \in B will get matched with at least one a \in A.
    The conclusion follows..
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  7. #7
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    It is almost impossible to read let alone follow your work.
    Why not learn to post in symbols? You can use LaTeX tags.
    [tex]f:A\to B[/tex] gives f:A\to B.

    Frankly, I donít understand your confusion with this.
    Moreover, I donít think you are justified in saying ďit is never said that a function has an inverse if and only if it is bijective. We remain silent about the if part.Ē You do not know that is the case.

    As a matter of definition: f:A\to B is a bijection if and only if
    f is injective and sujective.

    Now if f:A\to B is a bijection then  \left( {\forall t \in B} \right) define A_t  = f^{ - 1} \left( {\{ t\} } \right) = \left\{ {x \in A:f(x) = t} \right\}.
    Now it is completely obvious that each A_t is a nonempty singleton set, in fact they partition the set A.
    That means that there is a perfectly natural way to define g:B\to A as t\mapsto x\in A_t.
    The function g is a bijection and is the right and left inverse of f.
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