Problem 1.2.4: Let $X \subset \mathbb{R}^3$ be the union of $n$ lines through the origin. Compute $\pi_1 (\mathbb{R}^3 - X)$.
I have the following solution. Would someone be so kind as to check whether it's correct? Also, I have seen that there are solutions where one projects stereographically onto $\mathbb{R}^2$, but then people say that n holes in $\mathbb{R}^2$ is homotopy equivalent to $\lor_{n} S^1$ which they don't give any reasoning for. Would anyone know an argument for it?
My attempt:
Firstly take a small $\varepsilon$-neighborhood about $X$
and project all points in the compliment of this open set radially onto
$S^2$. The $\varepsilon$-neighborhood deformation retracts to $\mathbb{R}^3
- X$ and the projection is also a deformation retraction, so the
fundamental groups of the resulting space and $\mathbb{R}^3 - X$ are
isomorphic.
Our resulting space, call it $Y$, is a sphere with $2n$ disks removed, where $n$ is the
number of lines in $X$.
$Y$ is path-connected, so letting $x_0 \in Y$ be an arbitrary basepoint, we
can connect each boundary of the removed disks by a path in $Y$ to $x_0$.
We thus see that the resulting space can be constructed
as a CW-complex with $2n +1$ 0-cells (one for each circle that is the
boundary of the removed disk, and one for the chosen basepoint), $4n$
1-cells connecting, $2n$ used to construct circles and $2n$ used to
connected the circle to the basepoint. To this $1$-skeleton, we then
attach a $2-cell$ to get a sphere with $4$ disks removed. Since
$X$ is path-connected, we then find by proposition 1.26 that
$X^{1} \to X^{2}$ induces a surjection $\pi_1 \left( X^{1}, x_0 \right) \to
\pi_1 \left( X^{2}, x_0 \right) $
whose kernel is
generated by the attaching loop.
Now, the subcomplex consisting of the closure
of the edges that connect the circles to the basepoint
is contractible, so by page 11, the
quotient space obtained by collapsing this to a point
is homotopy
equivalent to the one skeleton, $X^{1}$.
And by proposition 1.18,
$\pi_1 (X^{1})$ is isomorphic to the fundamental group of the quotient
space which is $\pi_1 \left( \lor_{2n}S^{1} \right)
= *_{2n} \mathbb{Z}$. Therefore we get
$$
\pi_1 \left( X^2 \right)
\cong *_{2n} \mathbb{Z} / \left( a_1 a_2 \ldots a_{2n} \right) \cong
*_{2n-1} \mathbb{Z}
$$
where $a_i$ is the generating loop for
the $i$ th circle completed by the attaching loop,
giving $a_1 a_2 \ldots a_{2n}$ as the attaching loop
decomposed.
