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!

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How did you come up with your inverse? Did you actually just switch arguments? – k.stm Nov 1 '12 at 23:03
i actually did, i have a pretty poor math background..! – wwwe Nov 2 '12 at 1:02

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 $f^{-1}(a,b)$.
One should mention: Since $f$ is given by the matrix $\left(\begin{smallmatrix} 1 & 2 \\ 1 & -2 \end{smallmatrix}\right)$ of determinant $-4$, it isn't invertible as a endomorphism of $\mathbb{Z}^2$. This is also clear when looking at the second argument of the image: It's even. However, it must be invertible as a function, after all it's one-to-one. But this is more complicated to write down, you need to describe the image of $f$. – k.stm Nov 1 '12 at 23:20
True. In particular, it's not a bijection (you never get an odd number in the second entry), so the inverse will really only be defined on the image of $f$. – FPP Nov 1 '12 at 23:26