# Prove a bijection.

1. Oct 12, 2008

### 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 :)

2. Oct 12, 2008

### jjou

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

3. Oct 13, 2008

### estra

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

4. Oct 13, 2008

### HallsofIvy

Staff Emeritus
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.

5. Oct 13, 2008

### estra

OK. Lets 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)$$\in$$R 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$$\in$$R } R-1={(x,x)| x$$\in$$ R} and
R={(y,y)| y$$\in$$R } R-1={(y,y)| y$$\in$$R }

please somebody help me here :)
Thank you!

6. Oct 13, 2008

### estra

OK. Lets 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)$$\in$$R 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$$\in$$R } R-1={(x,x)| x$$\in$$ R} and
R={(y,y)| y$$\in$$R } R-1={(y,y)| y$$\in$$R }

please somebody help me here :)
Thank you!