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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 $\mathbb{C}P^{n+1}\setminus K$?

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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 $K$.

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    $\begingroup$ Would it be OK with you to edit in at the beginning: "No, $\mathbb C\mathbb P^n \setminus K$ is never a retract of $\mathbb C\mathbb P^{n + 1} \setminus K$ for $K$ nonempty"? $\endgroup$
    – LSpice
    Commented Feb 6, 2022 at 21:42
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    $\begingroup$ What about when $n=1$, and in particular when $K$ is a singleton? $\endgroup$
    – Tim Campion
    Commented Feb 7, 2022 at 1:37
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    $\begingroup$ @TimCampion I convinced myself that I checked the trivial cases but I guess this is not quite true. $\endgroup$
    – Will Sawin
    Commented Feb 7, 2022 at 1:44
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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?

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    $\begingroup$ By the way, obviously there is no retraction for $|K|>1$ since in that case $\mathbb{CP}^1\setminus K$ acquires a nontrivial fundamental group while $\mathbb{CP}^2\setminus K$ remains simply connected $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 7, 2022 at 13:49
  • $\begingroup$ Right, thanks. Together with Will Sawin's answer that seems to cover every case. $\endgroup$
    – Nick L
    Commented Feb 7, 2022 at 13:59

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