Let $I=[-1,1],$ and $J=\partial I=\{-1,1\},$ and consider $A=(I\times I\times J)\cup(I\times J\times I)$ ($A$ is a cube without two parallel faces). The fundamental group $\pi_1(C)$ is $\mathbb{Z}$ since $A$ deformation retracts to the square $\{0\}\times I\times I.$ I want to calculate $\pi_1(\mathbb{R}^3-A).$ Since I usually have problem with these kind of problems, I chose this one to make some questions. Are there some usual techniques to solve these problems: in particular based on van Kampen or deformation retracts? Are there some links between $\pi_1(X)$ and $\pi_1(Y-X)?$
Fundamental group of complement space
algebraic-topology
Best Answer
Let $S^2$ be a sphere of radius $2$ centred at the origin, and let $L$ be the straight line segment connecting the points $(2, 0, 0)$ and $(-2, 0, 0)$. The space $R^3 - A$ deformation retracts onto $S^2 \cup L$, whose fundamental group is $\mathbb Z$.
Hatcher, page 46, example 1.23 contains pictures of these kinds of deformation retractions for a number of examples, including $\mathbb R^3 - S^1$, which is very similar to the present example.