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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?

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    For one thing, the group that acts on $S^n_{\Bbb C}=S^{2n+1}$ is not $\Bbb Z/2\Bbb Z$, but $S^1=\Bbb R/\Bbb Z$: in $\Bbb C$ the relation on the unitary vectors is $x\sim y\iff \exists \theta\in\Bbb R,\ x=e^{i\theta}y$ and not "$x=\pm y$". – 2017-01-24
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    So the answer is: it is calculated in the same way, you just need to learn fancier stuff. – 2017-01-24

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The group that acts on $S^n_{\Bbb C}$ is not $\Bbb Z/2\Bbb Z$, but $S^1=\Bbb R/\Bbb Z$: in $\Bbb C$ the relation on the unitary vectors is $x\sim y\iff \exists \theta\in\Bbb R,\ x=e^{i\theta}y$ and not "$x=\pm y$".

The action of $S^1$ on $S^n_{\Bbb C}$ given by scalar complex multiplication, however, is not properly discontinuous and (therefore) the quotient map is not a covering.

A direct way to see that the standard projection $S^n_{\Bbb C}\stackrel\pi\longrightarrow \Bbb CP^n$ is not a covering is this: the fibers of a covering are by definition discrete sets. However, the fibers of this map are $$\pi^{-1}([v])=\{e^{i\theta}v\,:\,\theta\in\Bbb [0,2\pi)\}$$ which is homeomorphic to $S^1$.

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    (This is a fibration, though, and you can do "the same" thing one does with discrete fibrations = coverings.) – 2017-01-24
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    @PedroTamaroff I suspected as much, but, being still ignorant, I am not familiar with that theory. I was just pointing out where the proof he seemed to be suggesting fails. – 2017-01-24
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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