Let $X= S^1 \cup y$-axis in the $xy$-plane. Compute the fundamental group of $\mathbb{R}^3 – X$
Edit:
To clarify, $S^1 =\{(x,y)|x^2 + y^2 = 1\}$ is the unit circle in the $xy$-plane, so that the $y$-axis and the circle intersect at $(0,\pm 1).$
I tried doing this by first "blowing up" the circle and the line to get a homotopy equivalent space. It seems like when you do this blow up process you are left with a sphere with two holes in it and two line segments attached to the sphere at the endpoints of the line segment. The holes in the sphere come from $y$-axis and the line segments come from the space between the inside of the circle and the $y$-axis. Then, forgetting about the line segments for a moment we can project the rest of the space to $\mathbb{R}^2 – D^2$ via stereographic projection. The line segments are then arcs connecting two points in the plane (the arcs lying outside the plane). Finally the whole space deformation retracts to a wedge of 3 circles. Giving that $\pi_1(X) \cong \mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$.
This is a very visual process and I'm not sure my first space after "blowing up" is correct. Can anyone check this or give me suggestions. Also:
The space $X$ is not a knot, but can we still use the Wirtinger presentation to compute $\pi_1(X)$ by considering intersections as both over and under crossings?
Best Answer
Let $A=\{(x,y,z) \in X : x<\tfrac{1}{2}\}$ and $B=\{(x,y,z) \in X: x>-\tfrac{1}{2}\}$. Then $A$ and $B$ are homeomorphic and each deform retracts onto a surface $\Sigma$ surrounding "half" of the removed subspace; see below. Then $\pi_1(A)\cong \pi_1(B)\cong\pi_1(\Sigma)\cong \mathbb{Z} * \mathbb{Z} *\mathbb{Z}$. We can represent the three generators via paths as in the image below. This choice makes it clear that corresponding generators of $\Sigma_A$ and $\Sigma_B$ are equal when included into $A \cap B$. By van Kampen's Theorem, we have $\pi_1(X)\cong \mathbb{Z}* \mathbb{Z}*\mathbb{Z}$.