Local Isomorphism on Topological Groups I'm currently studying Lie Groups, and reading "Theory of Lie Groups I" by C. Chevalley. He talks about topological groups in chapter two. To be more precise, on page 38 he presents two examples in order to show that it's not always possible to extend a local isomorphism to the whole group. He says:
"Let $\phi$ be the mapping which assigns to every real number $x$ its residue class $\phi(x)$ modulo 1; let $f$ be the contraction of $\phi$ to the interval $]-1/4,+1/4[$. It is clear that f is a local isomorphism of $R$ into $T$, but these groups are not isomorphic."
Well, I've got a few doubts... Does it make sense to consider a residue class of a real number modulo 1? I don't understand that, since residue classes are taken to integers (sorry for my lack of knowledge if I'm wrong). Second, does "contraction" mean the same as in metric spaces? $T$ is the 1-dimensional torus (in general, the factor group $R^{n}/H$, where $H$ is the subgroup of $R^{n}$ where the elements have integers coordinates, is the $n$-dimensional torus $T^{n}$). Well, I didn't understand neither why this local isomorphism happens.
 A: I do not understand why the author use "contraction" (did Chevalley define it earlier? I could not find it in Google Books), it is not widely used in the modern Lie group literature. But suppose you are given with the local quotient map $$\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}$$ It is clear that at the open interval $(-\frac{1}{4},\frac{1}{4})$ the quotient map's image is homeomorphic to itself. Now the original additive group structure is also inherited from the quotient structure (since $\mathbb{Z}$ is also an additive subgroup). To be precise for any $x\in (-\frac{1}{4},\frac{1}{4})$, let $\epsilon\in (-\frac{1}{4},\frac{1}{4})$ be small enough such that $x+\epsilon\in (-\frac{1}{4},\frac{1}{4})$. Then we can claim $f(x)+f(\epsilon)\in f(-\frac{1}{4},\frac{1}{4})$ as well. So $f$ is well defined in terms of group structure. I guess restricting to an open neighborhood there exists a group isomorphism is what "locally isomorphic" means. 
A: Modulo 1 means taking the fractional part. If $x$ is a real number and $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$, then $x-\lfloor x\rfloor \equiv x \bmod 1.$ 
Because $im(\phi)$ is interval $(0,1)$ my guess is that the contraction $f:(0,1) \rightarrow (-1/4. 1/4)$. 
Here torus means the circle $S^{1} = \mathbb{R} / \mathbb{Z}$. 
