Can a punctured $\mathbb{C}P^n$ be a retract of a punctured $\mathbb{C}P^{n+1}$? In this question we consider the standard inclusion of $\mathbb{C}P^{n}\subset \mathbb{C}P^{n+1}$.
It is well known that  $\mathbb{C}P^n$ is  not a retract of  $\mathbb{C}P^{n+1}$.
What about  if  we remove a finite subset as  follows:
Question: Assume that $K\subset \mathbb{C}P^n$ is a finite set. Can $\mathbb{C}P^n\setminus K$ be  a  retract of  $\mathbb{C}P^{n+1}\setminus K$?
 A: No, $\mathbb{CP}^n \setminus K$ is never a retract of $\mathbb{CP}^{n+1} \setminus K$.
If $K$ is nonempty, the generator $x$ of  $H^2( \mathbb {CP}^n\setminus K )$  satisfies $x^n=0$. So its pullback along any map $\mathbb{CP}^{n+1} \setminus K \to \mathbb{CP}^n \setminus K$ must be a class $y$ satisfying $y^n=0$.
Since $H^i(\mathbb{CP}^{n+1} \setminus K ) \to H^i(\mathbb {CP}^{n+1})$ is an isomorphism for $i=2, 2n$, any such class must be $0$: It must be a pullback from $\mathbb {CP}^{n+1}$, and if it is nonzero then it is the pullback of a nonzero class so the $n$th power of the pullback is nonzero, thus the pullback of the $n$th power is nonzero, hence the $n$th power is nonzero.
But if $X$ is a retract of $Y$, then $X \to Y \to X$ is an isomorphism, so $H^i (X) \to H^i(Y)$ must be injective, so the pullback of $x$ cannot be zero. This is a contradiction, so it's not a retract.
However, $\mathbb {CP}^n$ is a retract of $\mathbb {CP}^{n+1} \setminus K$ for any nonempty $K$, by embedding as a hyperplane disjoint from $K$ and then projecting from a point in $K$.
A: I claim that for $n=1$ , $|K|=1$ there is such a retraction.
$\mathbb{CP}^2 \setminus K$ is the total space of the bundle $\mathcal{O}(1)$ over  $\mathbb{CP}^1$.
$\mathbb{CP}^1 \setminus K = \mathbb{C}$ is a fibre of this line bundle.
Then one may build such a retraction following the answer to thi question.
Is a topological fiber-bundle, whose total space admits a retraction onto a fiber, trivial?
