About the universal cover Let $X = S^1 \cup \{(x,y) : y = 0, x \neq 0\}$, that is $S^1 \cup \mathbb{R}\setminus \{0\}$. I'm triying to construct the universal cover of this space. This kind of space usually has something like the next: take $\mathbb{R}$ and put a vertical line in each integer. But this case is not possible now because there are a part of the line that is inside of the circle and this creates problems with the continuity. Any idea?
 A: The answer is more or less as you state in your post, although I might express it a bit more carefully: take $\mathbb R$ and attach a line at each integer and half-integer.
Here is a more precise description which takes advantage of a very nice construction that we already know, namely the exponential map $\exp : \mathbb C \to \mathbb C - \{0\}$.
Let $\widetilde X \subset \mathbb C$ be the imaginary axis together with the horizontal line at each integer and half-integer multiple of $2 \pi i$. More precisely, $\widetilde X$ is the union of the following pieces:

*

*The imaginary axis $\mathbb I = \{0+iy \mid y \in \mathbb R\}$

*For each integer $n \in \mathbb Z$, the line $\mathbb R + \pi n i = \{x + \pi n i \mid x  \in \mathbb R\}$
The exponential map $\exp : \mathbb C \mapsto \mathbb C - \{0\}$ restricts to a universal covering map from the subspace $\widetilde X \subset \mathbb C$ to your subspace $X \subset \mathbb C - \{0\}$. Note that  \begin{align*}
\exp(\mathbb I) &= S^1 \\
\exp(\mathbb R + \pi n i) &=
\begin{cases}
\text{the positive open half line of the real axis, if $n$ is even} \\
\text{the negative open half line of the real axis, if $n$ is odd}
\end{cases}
\end{align*}
