Fundamental Group of the complement of $\mathbb{S}^1$ union with $z$ axis in $\mathbb{R}^3$

Can you help me to compute the fundamental group of $\mathbb{R}^3\setminus X$, where $$X = \mathbb{S}^1\cup \{(0,0,z)\in\mathbb{R}^3\mid z\in\mathbb{R}\}\ ?$$

I'm sorry if I repeated the question and I'm sorry if the question it's not well written and ... thanks in advance ! I'd say that the fundamental group is isomorphic to $F_2$, but I'm not even sure if the intuition is right or not.

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What do you mean by $\mathbb{S}^1$? – Mercy King Jun 23 '12 at 19:56
It seems straightforward that your space is homotopy equivalent to the torus $\mathbb{S}^1\times\mathbb{S}^1$, so its fundamental group should be isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$. – Olivier Bégassat Jun 23 '12 at 19:56
See Example 1.23 in Hatcher. – Brandon Carter Jun 23 '12 at 20:00
I've never seen the blackboard S used for the circle, just the normal $S$. Is this standard? This notation confused me a little. – Potato Jun 23 '12 at 20:21
Actually, @thetruth, what exactly do you mean by $S^1$? I can think of many ways to imbed a circle in $\mathbb{R}^3$. – Potato Jun 23 '12 at 20:25

Thanks to S.V.K., you can add a point at infinity to obtain $S^3 - X$ where X are your now linked circles without changing the fundamental group. Then in a way similar to deformation retracting $\mathbb{R}^3 - S^1$ to obtain $S^2 \vee S^1$, you can deformation retract your new space to obtain $S^2 \vee T^2$, so that again by S.V.K. the fundamental group becomes $\mathbb{Z} \times \mathbb{Z}$.