Composition of two functions in $\mathbb{Z^2}\to \mathbb{Z^2}$ I need to find the composition of a function and its inverse so I have the identity function in return. My problem is that I don't seem to undestand how to proceed algebraically.
I have a function with couples, let's say: $f(x,y)=(x+y,\;2x-2y)$
The inverse would be: $f^{-1}(x,y)=(2x-2y,\;x+y)$
Now, to prove that $f^{-1}$ $\circ$ $f$ = identify function is where I'm getting confused.
I've been trying to solve this way:
$f^{-1}(f(x,y)) = 2(x+y)-2y,x+(2x-2y)$ because
I thought it was the way to proceed but it gives me an incorrect result.
Anyone could point me in the right direction?
Thanks!

Your inverse is not correct, and your calculation shows that.  To find the inverse, set $$(a,b)=f(x,y)=(x+y,2(x-y))$$ and solve for $x$ and $y$ in terms of $a$ and $b$.  That'll give your your inverse $f^{-1}(a,b)$.
