Supplementary Exercises*: Topological Groups: Problem 6 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

The integers $\mathbb{Z}$ are a normal subgroup of $(\mathbb{R},+)$ . The quotient $\mathbb{R}/\mathbb{Z}$ is a familiar topological group; what is it?

$[0,1)$ with addition by modulus, which is homeomorphic and isomorphic to $(S^{1},\cdot)$ . Indeed, $x\mathbb{Z}=y\mathbb{Z}$ iff $y=x+z$ for some $z\in\mathbb{Z}$ , so let $f(x)=e^{i2\pi x}$ , then $f$ is a quotient map and $f(x)=f(y)$ iff $y=x+z$ for some $z\in\mathbb{Z}$ iff $x\mathbb{Z}=y\mathbb{Z}$ , i.e., by Corollary 22.3, $f$ induces the homeomorphism $h:\mathbb{R}/\mathbb{Z}\rightarrow S^{1}$ (via the quotient $p:\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$ ) such that $h(x\mathbb{Z})=f(p^{-1}(x\mathbb{Z}))=f(\{x+z|z\in\mathbb{Z}\})=e^{i2\pi x}$ , in particular, $h(x\mathbb{Z}+y\mathbb{Z})=$ $h((x+y)\mathbb{Z})=$ $e^{i2\pi(x+y)}=h(x\mathbb{Z})\cdot h(y\mathbb{Z})$ .

Subpages

Supplementary Exercises*: Topological Groups: Problem 6 Solution

Supplementary Exercises*: Topological Groups: Problem 6 Solution