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Homework Help: Prove a bijection.

  1. Oct 12, 2008 #1
    Prove that R ⊂ X x Y is a bijection between the sets X and Y, when R−1R= I: X→X and RR-1=I: Y→Y

    Set theory is a quite a new lesson for me. So I am not good at proving different connections, but please give me a little help with what to start and so.. I have read the book and I know what bijection means but just..
    I would be very very thankful. Getting a good example proof or just some tips what to do would improve my skills also I hope :)

    Thank you in advance!
     
  2. jcsd
  3. Oct 12, 2008 #2
    How do you define [tex]R^{-1}[/tex] without assuming R is a bijection (and thus has an inverse)?
     
  4. Oct 13, 2008 #3
    I only know that if R ⊂ X x Y then R-1 ⊂ Y x X
    or: (y,x)[tex]\in[/tex]R-1 [tex]\Leftrightarrow[/tex] (x,y) [tex]\in[/tex] R
     
  5. Oct 13, 2008 #4

    HallsofIvy

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    Science Advisor

    But that is undefined for specific y if there is no pair with that y in the second position.

    And it is not uniquely defined if there is more than one pair with that y in the second position.
     
  6. Oct 13, 2008 #5
    OK. Lets assume that R [tex]\subset[/tex] X x Y is a bijection between X and Y.
    Then we also have to have R-1. What is R-1 ⊂ Y x X
    when i: X → X then (x,y)[tex]\in[/tex]R has to be written (f(x),x) [tex]\in[/tex] R and x [tex]\in[/tex] X and also (f-1)y,y [tex]\in[/tex] R and x [tex]\in[/tex] X.
    but how to show that R-1R=I abd RR-1=I ?
    or do we have to write R={(x,x)| x[tex]\in[/tex]R } R-1={(x,x)| x[tex]\in[/tex] R} and
    R={(y,y)| y[tex]\in[/tex]R } R-1={(y,y)| y[tex]\in[/tex]R }

    please somebody help me here :)
    Thank you!
     
  7. Oct 13, 2008 #6
    OK. Lets assume that R [tex]\subset[/tex] X x Y is a bijection between X and Y.
    Then we also have to have R-1. What is R-1 ⊂ Y x X
    when i: X → X then (x,y)[tex]\in[/tex]R has to be written (f(x),x) [tex]\in[/tex] R and x [tex]\in[/tex] X and also (f-1)y,y [tex]\in[/tex] R and x [tex]\in[/tex] X.
    but how to show that R-1R=I abd RR-1=I ?
    or do we have to write R={(x,x)| x[tex]\in[/tex]R } R-1={(x,x)| x[tex]\in[/tex] R} and
    R={(y,y)| y[tex]\in[/tex]R } R-1={(y,y)| y[tex]\in[/tex]R }

    please somebody help me here :)
    Thank you!
     
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