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?
Solution
Let \(\varphi\colon\mathbb R\to S^1\), \(\varphi(t)=e^{2\pi it}\).
1. \(\varphi\) is a continuous surjective homomorphism from \((\mathbb R,+)\) to \((S^1,\cdot)\) with kernel \(\mathbb Z\).
\(\varphi(s+t)=\varphi(s)\varphi(t)\), every point of \(S^1\) is \(e^{2\pi it}\) for some \(t\), and \(e^{2\pi it}=1\) if and only if \(t\in\mathbb Z\).
2. \(\varphi\) is an open map.
For each \(a\in\mathbb R\), \(\varphi\) maps \((a,a+1)\) homeomorphically onto the open arc \(S^1-\{\varphi(a)\}\), so it maps open subsets of \((a,a+1)\) to open subsets of \(S^1\); every open subset of \(\mathbb R\) is a union of such sets.
3. Q.E.D.
By step 1 and the first isomorphism theorem, \(\varphi\) induces a group isomorphism \(\mathbb R/\mathbb Z\to S^1\). By steps 1 and 2, \(\varphi\) is a quotient map, so this bijection is a homeomorphism. Hence \(\mathbb R/\mathbb Z\cong\boxed{S^1}\) as topological groups.