TITLE: This quotient map is a covering map
QUESTION [7 upvotes]: 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?
Am I right?
I'm a beginner in this subject, so I'm not sure if I solved it correctly
Thanks

REPLY [2 votes]: 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.