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My lecture notes state that a basic example of covering space is $(\mathbb{S}^n,\pi)$ where $\pi$ is the projection $\pi(p) = [p]$ to $\mathbb{RP}^n$ the projective space considered as the sphere with opposite points identified.

Next, I wonder what is the intuition behind this result. How does one visualize this covering? Let me explain myself: in the case of $\mathbb{R}$ and $\mathbb{S}^1$ it is easy to see that each loop of the fundamental group can be unrolled into a segment $[k,k+n]$. Is there a similar view in this case?

user1868607
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Response to first thing: No, it is the quotient map $$\pi: S^n \to S^n/\sim (i.e., \mathbb{R}P^n)$$ $$x \mapsto [x].$$ Response to second thing: I don't see how your example is an "intuition for being a covering map". But the fact analogous to your example is that every loop in the fundamental group of $\mathbb{R}P^n$ can be corresponded to half a great circle connecting two antipodal points and its concatenations.

However, I'd rather say that "the intuition" for being a covering map coincides with its proof: it is due to the fact that it is a quotient of a properly discontinuous action of a discrete group (namely, $\mathbb{Z}/2\mathbb{Z})$.

Aloizio Macedo
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$\mathbb{R}P^2$ is the quotient of $S^2$ endowed with the equivalent relation $xRy$ if and only if $x=y$, or$ x=-y$. So it is this quotient map. It is a cover since the restriction of the canonical projection $\pi:S^n\rightarrow \mathbb{R}^n$ to an open hemisphere is a diffeomorphism onto its image. It like so you can visualize this covering, since $x$ and $-x$ are in two different open hemispheres, so $\pi$ is a double covering map.

Tsemo Aristide
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    "It is a cover since the restriction of the canonical projection $\pi:S^n\rightarrow \mathbb{R}^n$ to an open hemisphere is a diffeomorphism onto its image." That is very misleading. Your reasoning applies mutatis mutandis to $f: (0,3\pi) \to S^1;$ $t \mapsto e^{it}$, which is not a covering map. – Aloizio Macedo Dec 02 '17 at 15:47
  • @AloizioMacedo sorry to revive this dead post, but his comment says "is a diffeomorphism onto its image", is your $f$ a diffeomorphism onto its image? from where I'm sitting it doesn't even look injective. have I missed something? –  Aug 09 '20 at 00:55
  • @Math $f$ isn't a cover, but the reasoning that is used, without further elaboration, would apply to conclude that it is. (Thus, the reasoning is faulty.) More explicitly, one could say "It is a cover since the restriction to small open intervals is a diffeomorphism onto its image", for example. – Aloizio Macedo Aug 09 '20 at 01:01