on which it is invertible. Since the mapping is a composition of an invertible linear map and a discontinuous map that maps open sets to open sets, both of which preserve Lebesgue measure, a map is non-invertible if and only if there is an open set of $(x,y)$ that has two or more preimages. Let $a,b,c,d$ be defined as previously.

\[lem1\] Any map (\[eqmap\]) with $ad-bc=1$ will be noninvertible on an open set if and only if there are $(K,L)\in{{\bf Z}}^2\setminus(0,0)$ integers such that $$|Kc-La|<1~\mbox{ and }~|Kd-Lb|<1.$$

[[**Proof**]{} ]{}Suppose that we have $(x,y)$ and $(u,v)\in[0,1)^2$ such that $(x,y)\neq (u,v)$ but $f(x,y)=f(u,v)$. Then there is $(K,L)\in {{\bf Z}}^2\setminus (0,0)$ such that $ax+by=au+bv+K$ and $cx+dy=cu+dv+L$. These hold if and only if $$a(x-u)+b(y-v)=K~\mbox{ and }~c(x-u)+d(y-v)=L.$$ Using the fact that $a,b,c,d$ are non-zero and $ad-bc=1$ this implies that this holds if and only if $$x-u=Kc-La~\mbox{ and }~y-v=Kd-Lb.$$ Thus the mapping is many-to-one if and only if there are $K$ and $L$ such that $$|Kc-La|<1~\mbox{ and }~|Kd-Lb|<1$$ and the result follows. [\
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It is quite possible that there are simultaneously many solutions to the inequalities of Lemma \[lem1\]; for example, near $A=-1$ and $\alpha$ large one can find arbitrarily large numbers of integers $(K,L)$ 