Hawaiian Earrings are covering space for shrinking wedge of circles 
for any $P=(x,y) \in \mathbb{R}^2$,  Let $C_{P,n}$ denote a circle centred at $\big(x+\frac1n, y \big)$ of radius $\frac1n$. Let $W_{P}$ denote the shrinking wedge of circle $\bigcup_{n \in \mathbb{Z}} W_{(0,0)} $..

Define the space
$$X=\{0\} \times \mathbb{R} \cup \big(\bigcup_{i \in \mathbb{Z}}  W_{(0,4i)} \big)$$.
Show that X is a covering space for $W_{(0,0)}$.
Hatcher claims that $X$ is a covering space for $W_{(0,0)}$ but he did not provide any hint. I tried some maps but I wonder if they will be covering space. I was thinking that the number of circles in $X$ are countable and this is the case with $W_{(0,0)}$. Hence there is a bijection between them. Hence a circle in $X$ would map to a circle in $W_{(0,0)}$, but then I doubt this would be continous. Please help me with this problems.
 A: You map the copy of $\mathbb{R}$ in the usual winding fashion onto the outermost circle. In your case, this should be $(0, y) \mapsto \left(\cos\left(\frac{\pi}{2}y\right), \sin\left(\frac{\pi}{2} y\right)\right)$. As for the copies of the earrings, you map them to the earrings in the base space after skipping the outer most circle. So, $C_{P, n} \subset W_{(0, 4i)}$ in the total space gets mapped to $C_{P, n+1}$ in the base.
A: Let us prove a slightly more general result:
Let $(X,x_0) \vee (Y,y_0)$ be the wedge (= one point union) of two pointed spaces, let $p : \tilde X \to X$ be a covering map and $F = p^{-1}(x_0)$. On the disjoint union $D =  \tilde X + (Y \times F)$ define an equivalence relation $\sim$ by identfying $\xi \in F \subset \tilde X$ with $(y_0,\xi) \in Y \times F$ for all $\xi \in F$. Let $\pi : D \to Z = D/\sim$ denote the quotient map. Intuitively $Z$ is obtained from $\tilde X$ by gluing at copy of $Y$ to each $\xi \in F$ via identifying $y_0$ and $\xi$.
Define $r : D \to (X,x_0) \vee (Y,y_0), r(\xi) = p(\xi)$ for $\xi \in \tilde X$ and $r(y,\xi) = y$ for $(y,\xi) \in Y \times F$. This induces a unique map $\rho : Z \to (X,x_0) \vee (Y,y_0)$ which is clearly surjective. We claim that it is a covering map.
Let $U$ be an open subset of $X$ which is evenly covered by $p$. Write
$p^{-1}(U) = \bigcup V_\alpha$ with pairwise disjoint open $V_\alpha \subset \tilde X$ which are mapped by $p$ homeomorphically onto $U$.
Now let $x \in X$.

*

*If $x$ has an open neigborhood $U$ evenly covered by $p$ such that $x_0 \notin U$, then $U$ is open in $(X,x_0) \vee (Y,y_0)$ and $\rho^{-1}(U) = \bigcup V_\alpha$ which shows that $U$ is evenly covered by $\rho$.


*If $x$ has an open neigborhood $U$ evenly covered by $p$ such that $x_0 \in U$, then $(U,x_0) \vee (Y,y_0)$ is open in $(X,x_0) \vee (Y,y_0)$ and $\rho^{-1}((U,x_0) \vee (Y,y_0)) = \bigcup (V_\alpha,\xi_\alpha) \vee (Y,y_0)$ which is the union of pairwise disjoint open subsets of $Z$ which are mapped by $\rho$ homeomorphically onto $(U,x_0) \vee (Y,y_0)$. Thus $(U,x_0) \vee (Y,y_0)$ is evenly covered by $\rho$.
Similarly, let $y \in Y$. Since $x_0$ has an open neigborhood $U$ evenly covered by $p$, the argument of 2. above shows that $(U,x_0) \vee (Y,y_0)$ is an open neigborhood of $y$ evenly covered by $\rho$.
Let us come to your question. Let $H$ denote the Hawaiian earring with basepoint $h_0$ = point where all circles meet. Then $(S^1,*) \vee (H,h_0)$ is homeomorphic to $H$ and our above construction with the standard covering map $p : \mathbb R \to S^1$ gives us the desired covering of the Hawaiian earring,
