Proving Bijection Between X and Y: Tips & Examples

[estra]

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!

 

[jjou]

How do you define R^{-1} without assuming R is a bijection (and thus has an inverse)?

 

[estra]

I only know that if R ⊂ X x Y then R^-1 ⊂ Y x X
or: (y,x)\inR^-1 \Leftrightarrow (x,y) \in R

 

[HallsofIvy]

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.

 

[estra]

OK. Let's assume that R \subset 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)\inR has to be written (f(x),x) \in R and x \in X and also (f^-1)y,y \in R and x \in X.
but how to show that R^-1R=I abd RR^-1=I ?
or do we have to write R={(x,x)| x\inR } R^-1={(x,x)| x\in R} and
R={(y,y)| y\inR } R^-1={(y,y)| y\inR }

please somebody help me here :)
Thank you!

 

[estra]

OK. Let's assume that R \subset 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)\inR has to be written (f(x),x) \in R and x \in X and also (f^-1)y,y \in R and x \in X.
but how to show that R^-1R=I abd RR^-1=I ?
or do we have to write R={(x,x)| x\inR } R^-1={(x,x)| x\in R} and
R={(y,y)| y\inR } R^-1={(y,y)| y\inR }

please somebody help me here :)
Thank you!