Let $n\geqslant 2$, I am asked to prove the following:
Proposition. $\pi_1(\mathbb{C}P^n,\cdot)$ is isomorphic to $\pi_1(\mathbb{C}P^{n-1},\cdot)$.
Proof. First, let us introduce some notation: $$p:=[1:0:\cdots:0]\in\mathbb{C}P^n,U_0:=\left\{[x_0:\cdots:x_n]\in\mathbb{C}P^n;x_0\neq 0\right\}.$$ Let define the following map: $$F:\left\{\begin{array}{ccc}[0,1]\times\mathbb{C}P^n\setminus\{p\}&\rightarrow&\mathbb{C}P^n\setminus\{p\}\\(t,[x_0:\cdots:x_n])&\mapsto&[(1-t)x_0:\cdots:x_n]\end{array}\right..$$ $F$ is well-defined and is a strong deformation retraction of $\mathbb{C}P^n\setminus\{p\}$ onto $\mathbb{C}P^n\setminus U_0$, therefore, one has: $$\pi_1\left(\mathbb{C}P^n,\cdot\right)\cong\pi_1\left(\mathbb{C}P^n,\cdot\right).\tag{1}$$ Besides, the following map is an homeomorphism: $$\left\{\begin{array}{ccc}\mathbb{C}P^n\setminus U_0\rightarrow\mathbb{C}P^{n-1}\\{}[x_0:x_1:\cdots:x_n]\mapsto[x_1:\cdots:x_n]\end{array}\right..$$ Therefore, one has: $$\pi_1\left(\mathbb{C}P^n\setminus U_0,\cdot\right)\cong\pi_1(\mathbb{C}P^{n-1},\cdot).\tag{2}$$ Finally, using $(1)$ and $(2)$, one gets: $$\pi_1\left(\mathbb{C}P^{n-1},\cdot\right)\cong\pi_1\left(\mathbb{C}P^n\setminus\{p\},\cdot\right).$$ $\Box$
I have trouble understanding how can I go on in the proof, especially why one has: $$\pi_1\left(\mathbb{C}P^{n}\setminus\{p\},\cdot\right)\cong\pi_1\left(\mathbb{C}P^n,\cdot\right)?$$ Any help will be greatly appreciated.
If $M$ is any connected manifold of dimension $n\ge 3$ then you can use Van Kampen's theorem to prove that removal of one point does not change the fundamental group: $\pi_1(M,p) \approx \pi_1(M-x,p)$ for any $x \ne p$.
Use Van Kampen's theorem with the open sets $U=M-x$ and $V=$any neighborhood of $x$ homeomorphic to an open ball of dimension $n$. Make use of the fact that $U \cap V$ is an open ball minus a point and is therefore homotopy equivalent to $S^{n-1}$ which is simply connected because $n-1 \ge 2$.