Construct the covering homomorphism and the covering groups of $S^1$ having the form $R/pZ$ where $pZ$ is the set of integral mutliples of the integer $p$.
This is the first part of an excercise from Lie Groups and Algebras with applications to Physics, Geometry and mechanics.
Since I'm doing a self study, I would appreciate feedback on my proof, regarding correctness, alternative shorter proofs and style.
A definition of covering groups can be found here: [covering groups][1].
Define $S^1 = R/Z$ with group operation $(+)$. Let $\pi$ and $\pi_p$ be the quotient maps from $R$ onto $R/Z$ and $R/pZ$.
Define $f: R/pZ \rightarrow R/Z$ by $f(\pi_p(x)) = \pi(x)$.
Then $f$ is well defined since $f(\pi_p(x+np)) = \pi(x+np) = \pi(x) + \pi(np) = \pi(x) + 0 = \pi(x)$ for any $n$ in $Z$.
Also $f$ is group homomorphism since $f(\pi_p(x) + \pi_p(y)) = f(\pi_p(x + y)) = \pi(x+y) = \pi(x)+\pi(y)$, where we use that quotient maps of topological quotient groups are group homomorphisms.
Now we prove that $f$ is also continuous. Let $U$ be any open subset of $R/Z$, then by definition of the quotient topology we have that $\pi^{-1}(U)$ is open in $R$. Note by definition of $f$ we have that $\pi = f \circ \pi_p$. It follows that $\pi^{-1}(U) = \pi_p^{-1}(f^{-1}(U))$. Therefore, also $\pi_p^{-1}(f^{-1}(U))$ is open in $R$. By the definition of the quotient topology $f^{-1}(U)$ is open. Hence $f$ is continuous.
To prove: $f$ is onto. We have that $\{f(\pi_p(x)) | x \in R\} = \{\pi(x)|x \in R\} = R/Z$.
To prove: $f$ is locally one to one. This proof will require some lemmas.
Lemma 1 let $x \in R$ than $x$ has a unique decomposition $x = a + b +cp$ where $a \in [0, 1)$, $b,c \in Z$ and $0 \le b \le p-1$. Stated without proof.
Lemma 2 let $x, x' \in R$. Let $x = a + b + cp$ and $x' = a'+ b'+ c'p$ be the unique decompositions of Lemma 1. Then $\pi(x) = \pi(x') \iff a = a'$. And $\pi_p(x) = \pi_p(x') \iff a = a'$ and $b = b'$. Stated without proof.
Lemma 3 let $x \in R$ then $f^{-1}(\pi(x)) = \{\pi_p(x),\ldots, \pi_p(x+p-1)\}$. Moreover, $|f^{-1}(\pi(x))| = p$.
Proof: \begin{align} f^{-1}(\pi(x)) =& \{\pi_p(x')|x'\in R, f(\pi_p(x')) = \pi(x)\}\\ =& \{\pi_p(x')|x'\in R, \pi(x') = \pi(x)\} &\text{ by def of $f$}\\ =& \{\pi_p(a + b' + c'p)|b',c' \in Z, 0 \le b \le p-1\} &\text{by lemma 2}\\ =& \{\pi_p(a + b')|b' \in Z, 0 \le b \le p-1\}&\text{(1.1)}\\ =&\{\pi_p(x),\ldots, \pi_p(x+p-1)\} \end{align} By applying lemma 2 on equation (1.1) we also find that $|f^{-1}(\pi(x))| = p$.
Using these lemma's we can now prove that $f$ is locally 1-1.
Let $x \in R$. Define $A = \{\pi_p(z)|z = x + \alpha, \alpha \in (-0.1,0.1)\}$. Then $A$ is an open neighberhood of $\pi_p(x)$.
We will prove that $\left.f\right|_A$ is 1-1. Obviously $\left.f\right|_A$ is surjective onto its image. we still need to show it is injective.
Let $y \in R$ be such that $\pi(y) \in f(A)$. We need to show that $|f^{-1}(\pi(y)) \cap A | = 1.$
Then $\pi(y) = \pi(x+\alpha)$ for some $\alpha \in (-0.1,0.1)$, by the definition of $f$ and the definition of $A$. By lemma 3 we have that $f^{-1}(\pi(y))=\{\pi_p(x+\alpha), \ldots, \pi_p(x+\alpha + p - 1)\}.$
To prove: $|f^{-1}(\pi(y)) \cap A | \ge 1.$
Proof: We have that $\pi_p(x+\alpha)$ is in $f^{-1}(\pi(y))$ and $A$.
To prove: $|f^{-1}(\pi(y)) \cap A | \le 1.$ For a contradition assume that $\pi_p(x+\alpha + d) \in A$ for some $d \in \{1, \ldots, p-1\}$. Then there is some $\alpha' \in (-0.1,0.1)$ such that $\pi_p(x+\alpha + d) = \pi_p(x+\alpha')$. So $x+\alpha + d = x+\alpha' + np, \text{ for some $n$ in Z} $
$\alpha - \alpha' = np - d$
$|\alpha - \alpha'| = |np - d|$
By the domains of $\alpha$ and $\alpha'$ we have that $|\alpha - \alpha'|\le 0.2$ and it can be shown that $|np - d|\ge 1$. Which gives a contradiction. Hence the only element of $f^{-1}(\pi(y))$ that is also in $A$ is $\pi_p(x+\alpha)$. Hence $f$ is locally 1-1.
We conclude that $R/pZ$ is a covering group with covering homomorphism $f$. [1]: https://en.wikipedia.org/wiki/Covering_group
What you have done is correct, but showing that $f$ is locally $1$-$1$ is not enough. It has to be proved that $f$ is a covering map which means that it is surjective (which is obvious from the definition) and that each $\xi \in \mathbb R / \mathbb Z$ has an open neigborhood $U$ such that $f^{-1}(U) = \bigcup_{\alpha \in A} V_\alpha$ with pairwise disjoint open $V_\alpha \subset \mathbb R / p\mathbb Z$ which are mapped by $f$ homeomorphically onto $U$. These $V_\alpha$ are called sheets over $U$. Here $A$ is a suitable index set.
Obviously we must have $\lvert A \rvert = \lvert f^{-1}(\xi') \rvert$ for all $\xi' \in U$. Since clearly all fibers of $f$ have $p$ elements, we therefore have to show that $f^{-1}(U)$ has a sheet decomposition of the form $\bigcup_{i=1}^p V_i$.
I suggest to proceed as follows:
Show that $\pi$ and $\pi_p$ are open maps. Note that $\pi = \pi_1$, so you do not need to consider two cases.
For $\xi = \pi(x)$ take $U = \pi((x-1/2,x+1/2))$. Verify that $V_i = \pi_p((x + 1/2 + i-1,x + 1/2 +i-1))$ gives the desired sheet decomposition of $f^{-1}(U)$.
By the way, note that $\mathbb R / p\mathbb Z$ is homeomorphic to the circle $S^1 = \{ z \in \mathbb C \mid \lvert z\rvert = 1\}$ for each $p$. A homeomorphism is given by $h_p : \mathbb R / p\mathbb Z \to S^1, h_p(\pi_p(x)) = e^{2\pi i x/p}$. Under these identifications $f$ corresponds to $\phi_p : S^1 \to S^1, \phi_p(z) = z^p$.