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Question about injectivity/surjectivity => bijectivity

  1. Oct 8, 2007 #1


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    we know that a bijective function is both injective and surjective

    injective means that unique values in the domain map to unique values in the codomain
    surjective means that all of the values in the codomain are mapped.

    but then, what of values in the domain that cannot be mapped to the codomain? (maybe we could take, for example, values of x < 0 for the function ln(x)).

    for example

    now, is Y -> X injective and surjective? Unique values within Y map to unique values within X, and all values of X are covered, if we count X as the codomain...
  2. jcsd
  3. Oct 8, 2007 #2


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    What do you mean by "Y->X"? It makes no sense to write "Y->X" without specifiying a function. Calling the function shown at that site "f", then f(1)= D, f(2)= B, and f(3)= A. The function is not surjective since nothing is mapped to C. That means that f does NOT have an inverse so you can't be talking about f-1:Y->X. Perhaps you mean the function, call it "g", such that g(D)= 1, g(B)= 2, g(A)= 3. But that, of course, is not from Y to X.
  4. Oct 8, 2007 #3

    matt grime

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  5. Oct 8, 2007 #4


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    okay, so I'm talking about say...

    domain: 1, 2, 3, 4
    codomain: 5,6,7

    f(x) : X-> Y s.t. f(1) = 5, f(2) = 6, f(3) = 7

    So here, we have that all the values in the codomain are mapped to a value in the domain (surjectivity) and that for all values in the domain, f(a) = f(b) iff a = b (which indicates injectivity). yet, what prevents this function from being bijective? (it clearly isn't 1-1 - it's only so if we restrict the domain), since f(4) cannot be mapped to anything).
  6. Oct 8, 2007 #5
    It's not a bijective function because it's not a function full stop. A function assigns an image to _every_ element of its domain. You haven't defined f(4), so you haven't defined a function.
  7. Oct 11, 2007 #6


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    only in algebraic geometry do we look at functions which are not defined everywhere on the source space. we call them rational maps to distinguish them from everywhere defined regular maps which we call morphisms.

    they usually arise as partial inverses of morphisms which are not injective or whose derivative is not everywhere injective.

    e.g. the morphism taking t to (t^2,t^3) is onto the curve y^2=x^3, but the inverse map taking (x,y) to y/x, is undefined at (0,0). we could of course set it equal to 0 there but it is not regular there anyway.

    if we consider the projection from the curve y^2 = x^3-x^2, to the y axis, taking a point of the curve to the place where the line joining that point to the origin meets the y axis, this projection is undefiend at (0,0), and in the limit sems to have two diferent klimiting values.

    i.e. the inverse map from the y axis to the curve taking a point of the y axis, to the point where the line joining the origin to the point f the y axis meets the curve, is defined everywhere on the y axis, but takes exactly two points of the y axis to the origin. anmely the two ponts where the y axis meets a tangent line to the curve at the origin (it has two tangent lines there).
  8. Oct 16, 2007 #7
    If a function is surjective "then all of the values in the codomain are mapped." What this statement also tells you is that the range of the function is precisely defined.

    If the function is also injective, then only one value in the codomain will be mapped to a given value in a domain, that will make the function bijective.

    So sometimes you have to restrict ranges of functions to make them bijective, like ln(x) for instance.
    Last edited: Oct 16, 2007
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