TITLE: Why is $\pi_1(\mathbb{C}P^n)$ not calculated in a similar way to $\pi_1(\mathbb{R}P^n)$?
QUESTION [3 upvotes]: Let $p:S^n\rightarrow \mathbb{R}P^n(n\geq 2)$ be the double covering. $S^n$ is simply connected and locally path connected, and $\mathbb{Z}/2\mathbb{Z}$ acts on this evenly, so $\pi_1(\mathbb{R}P^n)=\mathbb{Z}/2\mathbb{Z}$.
I think the assumption is satisfied in the complex case - complex sphere $S^n_{\mathbb{C}}$ and $\mathbb{C}P^n$, so $\pi_1(\mathbb{C}P^n)=\mathbb{Z}/2\mathbb{Z}.$ Where am I wrong?

REPLY [1 votes]: They can be computed in similar ways. Both of these results follow from the long exact sequence of a fibration. As G. Sassatelli pointed out it is $S^1$ that acts on $S^{2n+1}$, and the quotient of $S^{2n+1}$ by this action gives $\mathbb{CP}^n = S^{2n+1}/x\sim \lambda x$. We have two fibrations:
$$\mathbb{Z}/2\mathbb{Z}\xrightarrow{\ \ \ }S^n\xrightarrow{\ p\ }\mathbb{RP}^n\quad\text{and}\quad S^1\xrightarrow{\ \ \ }S^{2n+1}\xrightarrow{\ p \ }\mathbb{CP}^n.$$
These give us the long exact sequences:
$$\cdots\xrightarrow{\ \ \ }\pi_1(\mathbb{Z}/2\mathbb{Z},+1)\xrightarrow{\ \ \ }\pi_1(S^n,s_0)\xrightarrow{\ \ \ }\pi_1(\mathbb{RP}^n,\ast)\xrightarrow{\ \ \ }\pi_0(\mathbb{Z}/2\mathbb{Z},+1)\xrightarrow{\ \ \ }\pi_0(S^n,s_0)\xrightarrow{\ \ \ }\cdots$$
and
$$\cdots\xrightarrow{\ \ \ }\pi_1(S^1,s_0)\xrightarrow{\ \ \ }\pi_1(S^{2n+1},s_0)\xrightarrow{\ \ \ }\pi_1(\mathbb{CP}^n,\ast)\xrightarrow{\ \ \ }\pi_0(S^1,s_0)\xrightarrow{\ \ \ }\pi_0(S^{2n+1},s_0)\xrightarrow{\ \ \ }\cdots$$
Now, using the fact that $\pi_j(S^n)=0$ for $j<n$, we see that $\pi_1(\mathbb{RP}^n,\ast)\cong\pi_0(\mathbb{Z}/2\mathbb{Z},+1)\cong\mathbb{Z}/2\mathbb{Z}$ and $\pi_1(\mathbb{CP}^n,\ast)\cong 0$.
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