Proof of Isomorphism from $\mathbb R/\mathbb Z$ to $\mathbb R/2\pi\mathbb Z$ How can I construct a function $f$ to prove such isomorphism between $\mathbb R/\mathbb Z$ and $\mathbb R/2\pi\mathbb Z$?
The obvious one is to define $f: \mathbb R/\mathbb Z \to \mathbb R/2\pi\mathbb Z$ by $f(a+\mathbb Z)=2\pi a+2\pi\mathbb Z$
Are there any other good options for defining such a function so that it is simple to prove that it is a bijective homomorphism?
 A: One strategy is the following: 
Define an isomorphism $\varphi:\mathbb{R}\to \mathbb{R}$ which sends $\mathbb{Z}$ to $2\pi \mathbb{Z}$. Then, you can show that in general if $f:A\to A$ is an automorphism of an Abelian group taking a subgroup $N$ to $N'$, then $f$ induces an isomorphism $A/N\to A/N'$.
A: Intuitively, they should be isomorphic since they only differ by stretching with the factor $2\pi$. If you want to construct isomorphisms, the first isomorphism theorem is often helpful. Putting these together, we may want to look at the map $f\colon\mathbb{R}\rightarrow\mathbb{R}/2\pi\mathbb{Z},\,x\mapsto[2\pi x]_{2\pi\mathbb{Z}}$, where $[\cdot]_{2\pi\mathbb{Z}}$ denotes the equivalence class in $\mathbb{R}/2\pi\mathbb{Z}$ (i.e., $f$ is a stretching by $2\pi$ followed by projecting down). I leave it as exercise to show that $f$ is surjective and that $\ker f=\mathbb{Z}$, so, by the first isomorphism theorem, this map descends to an isomorphism $\tilde{f}\colon\mathbb{R}/\mathbb{Z}\rightarrow\mathbb{R}/2\pi\mathbb{Z}$, which is the same map as the one you have given in the post. The advantage of this method is that you get both well-definedness and bijectivity for free.
