Fundamental group of $\mathbb{R}^3 - X$

Question

Free group of $\mathbb{R}^3 - X$ where $X = \{ (x,y,0) \in \mathbb{R}^3 : x^2 + y^2 =1\} \cup \{(0,0,z) \in \mathbb{R}^3\}$...

Thus our space is 3 dimensional euclidean space with a planar circle at the origin and z-axis removed.

I was thinking that this space would be homotopy equivalent to the same space if we sort of bent the z-axis around on itself, I mean in that direction anyway, you couldn't close it because it is infinite, although perhaps you could take the one point compactification or something, but I don't know if that preserves homotopy type.

Anyway, thought like the one above, and nothing more rigorous than that, led me to think this space would have the same fundamental group as two un-knots with linking number one, or two circles interlocked, and thus I believe that the free group is $\mathbb{Z} * \mathbb{Z}$ but i'm not positive.

I think there is some complexity going on in the fact that I am trying to think about the z-axis as acting like one of the linked unknots, but since the z-axis is what is MISSING from our space i'm not sure if you can think of deforming the z-axis to bend like that and consider the homotopy type to be the same as before you bent it.

Any insight into my thinking and how it can be improved would be great!

this has been asked several times on this site. See here for pictures. — Andres Mejia, May 06 '19 at 21:46
Andres Mejia's link provides the picture, but the theoretical justification is this: By one-point compactifying you can see this is the same as $S^3$ minus two linked circles. In fact, if you consider the solid torus decomposition of $S^3$ we are removing the two core circles, so the space deformation retracts onto their common boundary which is $S^1 \times S^1$. — William, May 07 '19 at 11:41

Fundamental group of $\mathbb{R}^3 - X$

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