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Prove: A mapping f:S->T is bijective if and only if it has an inverse?

  1. Jun 18, 2004 #1
    Can a mapping from f:S->T associate an element of s into several elements of T? Also, how do you prove: A mapping f:S->T is bijective if and only if it has an inverse?
  2. jcsd
  3. Jun 18, 2004 #2


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    No. "mapping" is another word for "function". Part of the definition of function is that f(x) be a single value.

    f:S-> T is bijective iplies f has an inverse:

    Let y be any member of T. Since f is bijective it is both one-to-one (injective) and onto (surjective). Since f is surjective, there exist at least one member of S, x, such that f(x)= y. Since f is injective, there is only one such. Define g: T->S by g(y)= x.
    Now show that that is a function and is the inverse of f.

    f has an inverse implies f is bijective.

    i) for any y in T, let x= f-1(y). Then, by definition of inverse, f(x)= y so f is surjective.

    ii) suppose f(x1)= f(x2). Applying f-1 to both sides of the equation f-1(f(x1)= f-1(x2)
    x1= x2
    so f is injective.

    Since f is both injective and surjective, it is bijective.
  4. Jun 19, 2004 #3
    To take things into an account, I remember in discrete maths - a bit like computational mathematics or rather, it is computational mathematics - we define set |R| = |{1,1,1,1}| as 4. As there are four items in the set. Now, in the other extreme, let S = {1}, then |S| = |{1}| = 1. So in effect the sets R and S have different properties.

    One other thing is that, once again, in computer science, R = {1,1,1,1} and S = {1} are two totally different sets. Lets map (+2) to both sides and find the total sum of all the elements in the sets, i.e., {sum (map (+2) R)} and {sum (map (+2) S)}.

    sum (map (+2) R) = sum {3,3,3,3} = 12
    sum (map (+2) S) = sum {1} = 1 /= 12 = sum (map (+2) R)
    and as function applied two both sets are equal, R and S are not equal.

    However, answering your question, even in computer science the values of each element after mapping only takes 1 value from the image. i.e., notice that (map (+2) R) = {3,3,3,3}. So 1 -> 3 by mapping. And each element valued 1, goes to exactly one value in the image of f, that is 3.

    But as I said, this is in computer science and most of the computer applications - although duplicates are considered nasty in some cases.
    Last edited: Jun 19, 2004
  5. Jun 19, 2004 #4

    matt grime

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    The set {1,1,1,1} contains only one element. you aren't allowed to repeat elements in a set.
  6. Jun 21, 2004 #5
    I meant cardinalities... yes my wordings were pretty poor.
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