What would be the Fundamental Group of the complement of $k$ disjoint lines in $\mathbb{R}^3$.
If all the $k$ lines passes through the origin, then I know that it is the free group on $2k-1$ generators. If the lines are disjoint and does not pass through the origin, then does the Fundamental Group change? If it is same then how can I show that they are same? Please help!!
You can smoothly and nonintersectingly deform any finite set of nonintersecting lines in $\mathbb{R}^3$ so that they are all vertical. Your space is now clearly homeomorphic to $(\mathbb{R}^2 \smallsetminus \{p_1, \ldots, p_k\}) \times \mathbb{R}$.
Since $\mathbb{R}$ is nullhomotopic, your space is homotopic to the space $\mathbb{R}^2 \smallsetminus \{p_1, \ldots, p_k\}$, which has the free group $F_k$ as fundamental group.