Universal covering of the complement of a circle in $\mathbb{R}^3$ What is the universal covering of $X=\mathbb{R}^3\setminus(S^1\times\{0\})$?
I've been trying to build a covering map from $\mathbb{R}^3$ onto $X$ via composition of $p:\mathbb{R}^3\to Y$ and $q:Y\to X$ where $Y=\mathbb{R}^3\setminus\{(x,0,0):x\in\mathbb{R}\}$, $p(x,y,z)=(x,e^y\cos(z),e^y\sin(z))$ and $q$ is to be found, such that $(Y,q)$ is a covering space of $X$. 
Since $(\mathbb{R}^3,p)$ is the universal covering of $Y$, if I found a suitable $q$ then I hope that proving that $(\mathbb{R}^3,q\circ p)$ is the universal covering space of $X$ is not a pain. 
Is this a good approach? Is there any better solution?
 A: Your space $X$ is homotopy equivalent to $S^1 \vee S^2$. If $X$ and $Y$ are homotopy equivalent, then their universal covers are homotopy equivalent. The universal cover of $S^2 \vee S^1$ is $\Bbb R$ with a sphere wedged at each integer. So the universal cover of $X$ is homotopy equivalent to that. 
In particular, this is not homotopy equivalent to $\Bbb R^3$ - some advanced tools show $\pi_2(\tilde{X}) \cong \Bbb Z[t,t^{-1}]$ (example 4.27 in Hatcher). So $\tilde X$ couldn't possibly have been homeomorphic to $\Bbb R^3$, which has $\pi_2(\Bbb R^3) = 0$. I doubt you will find a nice way to write down a space homeomorphic to $\tilde X$; even the universal cover of $S^2 \vee S^1$ above is a bit messy.
Standard in knot theory is to consider the complement of your knot $K$ in $S^3$ rather than in $\Bbb R^3$. Indeed, $S^3 \setminus \{(x,y,0,0) : x^2 + y^2 = 1\}$ is homeomorphic to $\Bbb R^2 \times S^1$, so its universal cover is $\Bbb R^3$. Knot complements are always much better behaved in this setting; for instance, the universal cover of a knot complement in $S^3$ is always contractible.
