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What is is neither injective, surjective, and bijective?

  1. Oct 1, 2014 #1
    As the title says.
     
  2. jcsd
  3. Oct 1, 2014 #2

    Nugatory

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    Staff: Mentor

    Are you asking what term we would use to describe a mapping that is neither injective, surjective, nor bijective? Or are you asking for an example of suchba mapping?
     
  4. Oct 1, 2014 #3
    The criteria for bijection is that the set has to be both injective and surjective.
    In case of injection for a set, for example, f:X -> Y, there will exist an origin for any given Y such that f-1:Y -> X.
    In case of Surjection, there will be one and only one origin for every Y in that set. For example y = x2 is not a surjection.
     
  5. Oct 1, 2014 #4
    Yes sir, exactly.
     
  6. Oct 1, 2014 #5

    Char. Limit

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    [tex]x^2 + y^2 = 1[/tex]
     
  7. Oct 1, 2014 #6

    WWGD

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    To be more precise, as nuuskur pointed out, the function ## f : \mathbb R \rightarrow \mathbb R ## defined by ## f(x)= x^2 ## is neither injective nor surjective; f(x)=f(-x) , and no negative number is the image of any number. You need to clearly state your domain and codomain, otherwise every function is trivially surjective onto its image. If you changed/restricted the domain, OTOH, you can make the same _expression_ ##f(x)=x^2 ## a bijection from the positive Reals to themselves. While long-winded, the point is that you must define your domain and codomain in order to define a function and study its properties unambiguously.
     
  8. Oct 1, 2014 #7
    Yes, thank you WWGD. I apologize for my lazy explanation.
     
  9. Oct 2, 2014 #8
    Thank you sir
     
  10. Oct 2, 2014 #9

    WWGD

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    Hey, no problem , we all do it at times.

    No problem; it is the nitty-gritty, but it is necessary to do it at least once .
     
  11. Oct 5, 2014 #10
    I made a mistake. I swapped the terms. Any injection has only one origin for every Y. A surjection has at least one origin for every Y.
     
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