1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Prove a bijection.
  1. Proving Bijections (Replies: 1)

  2. Proving bijection (Replies: 4)

Loading...