Complement of a neighborhood of a smooth conic in $\Bbb CP^2$ In p.4 of this pdf: https://www.intlpress.com/site/pub/files/_fulltext/journals/pamq/2008/0004/0002/PAMQ-2008-0004-0002-a001.pdf, there is the following paragraph and I have some questions about it.

Let $C \subset \Bbb CP^2$ be a smooth conic and $C \subset N$ a regular neighborhood with boundary $L$. Since the normal bundle of $C$ in $\Bbb CP^2$ has degree $4$, we see that $L$ is a $\Bbb Z/4$-quotient of $S^3$. Set $M := \Bbb CP^2 \setminus \text{Int} N$. $M$ is a rational homology ball with $π_1(M) = \Bbb Z/2$ which bounds $L$.


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*I know that $C$ is homeomorphic to $S^2$ and that the normal bundle of $C$ in $\Bbb CP^2$ has degree 4, but why is $L$ a $\Bbb Z/4$ quotient of $S^3$?


*How do we know that $M$ is a rational homology ball with $\pi_1(M)=\Bbb Z/2$?
 A: Before answering let me recall the construction of the tautological line bundle over $\mathbb{CP}^1$ as $(p,q) \in \mathbb{CP}^1 \times \mathbb{C}^2\setminus \{0\}$ such that $\pi(q)=p$ where $\pi$ is the quotient map forming projective space.
1.
The key fact is that the line bundle $\mathcal{O}(4)$ over $\mathbb{CP}^1$ (resp. $\mathcal{O}(k)$ )  may be constructed by quotienting the tautological bundle by an action of $\mathbb{Z}_{4}$  (resp. $\mathbb{Z}_{k}$) on $\mathcal{O}(1)$ via bundle morphisms. Furthermore the action is free on the sphere bundle of this bundle. It is not hard to see that quotient will then be the sphere bundle of $\mathcal{O}(k)$.
In co-ordinate the action takes the form $(p,q) \mapsto (\xi p, \xi q)$ where $\xi$ is a 4th (resp $k$th) root of unity.
2.
One may conclude using Mayer-Vietoris. If $A$ is a neighbourhood of the conic and $M = B$ is a neighbourhood of the complement $\mathbb{CP}^2 \setminus A$. We that $A \cap B$ is homotopy equivelant to the above Lens space $S^3/\mathbb{Z}_4$, and $A \cup B = \mathbb{CP}^2$.
Looking at the MV sequence, $M$ is a rational homology ball since $H_{3}(A \cap B, \mathbb{Z}) = H_{4} (A \cup B, \mathbb{Z}) = \mathbb{Z}$  "cancel out", and $H_{2}(A , \mathbb{Z}) = H_{2} (A \cup B , \mathbb{Z}) = \mathbb{Z}$ "cancel out"  and there is nothing left to give rational homology in $M$.
Next we will show that $H_{1} (M) = \mathbb{Z}_{2}$ by using MV with a little care. It is known that for Lens space $X$ have $H_{2}(X,\mathbb{Z})=0$ (see here http://www.map.mpim-bonn.mpg.de/Lens_spaces or use the Universal coefficient theorem), so we have $H_2(A \cap B, \mathbb{Z})=0$ so continuing in the MV sequence from there we obtain:
$$ 0 \rightarrow \mathbb{Z} \oplus H_{2}(B,\mathbb{Z})  \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}_4 \rightarrow H_{1}(B , \mathbb{Z}) \rightarrow 0$$
From this we get easily that $H_{2}(B,\mathbb{Z}) = 0 $  so we modify.
$$ 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}_4 \rightarrow H_{1}(B , \mathbb{Z}) \rightarrow 0$$
Finally I claim that the inclusion $A \rightarrow A \cup B  = \mathbb{CP}^2$ induces the map $n \mapsto 2n$ in $H_2$. This is essentially the fact (i.e. Bezouts theorem) that if $C$ is a conic and $L$ is a line then there is the intersection product $$C.L=2.$$ So a conic represents twice the generator of $H_{2}(\mathbb{CP}^2, \mathbb{Z}) = \mathbb{Z}$.
Hence, the image of the map $\mathbb{Z} \rightarrow \mathbb{Z}_4$ must have exactly two elements. Finally the only possibility is that $\rightarrow H_{1}(B,\mathbb{Z})$ has two elements. Giving $ H_{1}(B,\mathbb{Z}) \cong \mathbb{Z}_2$.
Note that this only proves that the abelianisation of $\pi_1$ is $\mathbb{Z}_2$. To finish off we may use the Seifert-Van Kampen theorem (using the same $A,B$ as above). Let $J : A \cap B \rightarrow B$ be the inclusion.
Since $\pi_1(A)=\{1\}$(since it is a disk bundle over $S^2$), we have that $\{1\} = \pi_{1}(\mathbb{CP}^2) = \pi_{1}(B)/J_{*}(\pi_{1}(A \cap B)) = \pi_{1}(B)/J_{*}(\mathbb{Z}_{4})$.
From this we see that $\pi_{1}(B)$ is finite and furthermore $|\pi_{1}(B)| \leq 4.$ So using the fact that the abelianisation is $\mathbb{Z}_2$  and that the classifcation of groups with few elements completes the proof.
