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For any integer $n\ge 1$ the map $q:\mathbb S^n\to\mathbb {RP^n}$, which identifies antipodal points, is a covering map.

I'm trying to solve this question in the following manner (with the help of the comments and answers below):

Let $y$ be a point in $\mathbb {RP^n}$, and take any neighborhood $U$ of $y$. The preimage of $U$ are open subsets $V$ and $-V$, a question emerges, $q|V$ and $q|-V$ are homeomorphic to $U$? and why? if it does so, then we're done?enter image description here

Am I right?

I'm a beginner in this subject, so I'm not sure if I solved it correctly

Thanks

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    @user42912: I'm beginning to learn topology and have the same question. Have you received a satisfying answer since asking this question?2018-11-15

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Edit: This was written relating to an earlier, much different phrasing of the problem. That said:

This doesn't work as is. You need to look at the pre-image of y. It will never lie entirely in the set $U$ you defined as it contains antipodal points (just draw a picture in 3 dimensions - it contains 2 opposing points). This is also the key insight to show that $p$ is, in fact, a covering map. Hint:

(1) show it is open (calculate $p^{-1}(p(U))$ for some open set $U$ in $S^n$. Use how the antipodal map $a$ operates.)

(2) map back to a point $x$ in $p^{-1}(y)$. Visualize that in 3 dimensions, you can choose a neighborhood $U$ of $x$ small enough that it contains no antipodal points. Do the same for $n$. Then p is bijective from $U$ onto $p(U)$. Argue it is a homeomorphism (using (i); this is similar to what you do above). Do the same for the antipodal set $a(U)$. Conclude.

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    yes, my mistake, I'm pretty new here, I'm sorry. Thank you for you answer, it helped$a$lot.2012-12-13