Universal covering space of $\mathbb{R^3}\setminus S^1$

Question

I am having a problems with describing the universal covering space of $\mathbb{R^3}\setminus S^1$. Namely, I need to find a CW-space which it is homotopy equivalent to. Well, I don't know how to guess the space hence I cannot answer which CW-space it is equivalent to. I only know that it has infinite many fibers because $\pi_1(\mathbb{R^3}\setminus S^1) = \mathbb{Z}$.

Answer 1

Hint: consider countably many copies of $\Bbb R^3\setminus \operatorname{Int}(D^2)$ where $\partial D^2=S^1$. Glue all the $\operatorname{Int}(D^2)$ back in a way that they act as portals to lift from the $i$-th copy to the $i+1$.

Think of how the universal cover of the circle can be thought of as infinitely many circles with a point removed, glued back together in a way that the removed point is a lift to the next storey.

Answer 2

Here is a way to think about your space: $\mathbb{R}^3-S^1$ deformation retracts to a 3-dimensional ball minus $S^1$. As you mentioned, your the fundemental group of your space is isomorphic to $\mathbb{Z}$ and is generated by going once inside the removed $S^1$. With this in mind, most of our 3-dimensional ball is not needed since this $D^3-S^1$ deformation retracts to simply $S^2$ with its north and south poles connected by a line. This is the space you want to work with.

Answer 3

If $S^1$ is embedded in $\mathbb{R}^3$ as the unit circle in some plane (as opposed to a non-trivial knot), then $\mathbb{R}^3\setminus S^1$ deformation retracts to $S^2$ with a line between the north and south pole. This space is homotopy equivalent to $S^2\vee S^1$ which has universal cover $\mathbb{R}$ with an $S^2$ attached at every integer. This in turn is homotopy equivalent to a wedge of a countable number of copies of $S^2$.