I'd like to compute the fundamental group of $\mathbb{C}\mathbb{P}^{n}$ possibly using Van Kampen theorem, there is another source on SE which is Is complex projective space simply connected? but it goes beyond my actual knowledge.
It seems there is a lot of difference between the computation of the real and the complex projective fundamental group without using homology, since I didn't have found any material on classical books either.
What I'd like to do is to proceed by induction since I already know that $\mathbb{C}\mathbb{P}^{1} \sim \mathbb{S}^{2}$ simply connected. For the inductive step I'd like to define $A = \mathbb{C}\mathbb{P}^{n}- H$, where $H = \left\lbrace [x_{0} : \cdots : x_{n}] : x_{0} = 0\right\rbrace$ and $B = \mathbb{C}\mathbb{P}^{n}-[1 : \cdots : 0]$
I know that $A$ is homeomorphic to $\mathbb{C}^{n}$ so it has trivial fundamental group. I'd like to prove that $H$ is a deformation retract of $B$ to conclude that it's simply connected as well and conclude since the intersection is path connected.
I don't really how to properly (i.e formally) costruct the deformation on $H$. There are simple ways or nicer way to do it ?
Any help, hint or solution which doesn't require any theory greter than general topology and using Van Kampen would be appreciated.
Best Answer
Perhaps this helps. You can obtain $\mathbb CP^n$ from $\mathbb CP^{n-1}$ by gluing a disk of dimension $2n$ using the canonical map $S^{2n-1}\to \mathbb CP^{n-1}$. Hence, let us consider the situation that $X$ is obtained from $Y$ as $Y\cup_f e^{2n}$, starting at $n=1$. The claim is then that if $X$ is simply connected then so is $Y$. In fact, it should follow that the map $Y\to X$ where the cell is added induces an isomorphism on $\pi_1$.
To see this, consider a loop on $X$. Argue (by compactness or otherwise) that the loop can be assumed to miss at least one point in the interior of $e^{2n}$. Once this is done, you can use the fact removing a point from $e^{2n}$ allows you to retract it into $Y$, and the loop in this space is contractible.
Add. The map $S^{2n+1}\to \mathbb CP^n$ can also be used a bit more directly here. It shows that $ \mathbb CP^n$ is the quotient of $S^{2n+1}$ by the action of $S^1$. In fact, this projection is a fibration, and you have a long exact sequence of homotopy groups that gives you a precise description of the homotopy groups of $ \mathbb CP^n$. See here.