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I know that $\pi_1(CP^n)=0$ here is a possible proof:

Notation: for a CW complex, denote by $X^k$ the $k$-skeleton of $X$.

I will show that $\pi_1(CP^n)$ is contained in $\pi_1(S^2)=0$. Let $f:S^1 \longrightarrow CP^n$ be a map. By cellular approximation, $f$ can be assumed to be cellular. It follows that the image of $f$ lies in the $1$-skeleton of $CP^n$ that is: $f(S^1)\subset(CP^n)^1=S^2$. The statement follows.

Is this correct? Is there a more elementary way to show the result (cellular approximation isn't so trivial ) ? I think the long exact sequence of the fibration $S^1 \longrightarrow S^{2n+1} \longrightarrow CP^n $ only allows one to compute easily the $\pi_i$ for $i>2$.

Anthony Conway
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    http://math.stackexchange.com/questions/589134/fundamental-group-obtained-by-attaching-a-n-cell-with-n-2 – Chris Gerig May 29 '14 at 07:54
  • Yes, that's correct. An easier argument is to notice that every smooth loop in $CP^n$ is homotopic (by Sard's theorem) to a one which is disjoint from $CP^{n-1}$, i.e. lies in $C^n$, which is contractible. – Moishe Kohan May 29 '14 at 15:42
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    To elaborate on Chris Gerig's comment: One can prove, via more elementary methods, that the fundamental group depends only on the $2$-skeleton of a CW complex. Since $S^2$ is simply-connected, it follows that $\Bbb CP^n$ is simply-connected for all $n$. – Ayman Hourieh May 30 '14 at 10:07
  • @MoisheKohan can you elaborate on how you can apply Sard's theorem? The same argument wouldn't work for a map $S^2\to CP^n$, since $\pi_2(CP^n)=\mathbb Z$ for $n\ge1$. – Kenta S Apr 14 '22 at 20:38
  • @KentaS Suppose the image of the circle is transversal to $CP^{n-1}$. What can you say about their intersection? Can it be nonempty? – Moishe Kohan Apr 15 '22 at 00:13

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