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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


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I'm not sure if I follow your logic. Could you explain it a little bit more? Do you visually see why this is true (say for $n=2$)? – Alex Youcis Dec 11 '12 at 1:35
This is much better than the original post, and shows what is going on. "are we done?": you also need $V \cap -V = \emptyset$. For this, see (2) below: an intersection of a small open ball with $S^n$ is open in $S^n$. You know radius of $S^n$, and cam so easily choose a $V$ small enough (what radius of the ball will do?). To see the two restrictions are homeomorphisms, use (1) and the rest of (2) below. If still stuck, might add later, but busy now. – gnometorule Dec 13 '12 at 16:31
Why is (1) useful? If $p$ is a bijective continuous function that maps open sets to open sets, then its inverse maps open sets to open sets, and is so continuous too. – gnometorule Dec 13 '12 at 16:39
up vote 2 down vote accepted

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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I suggest to keep downvoting an answer outlining the correct solution. For similar reasons, I stopped participating in stackexchange proper. It can turn very YouTube. – gnometorule Dec 13 '12 at 2:14
I think you're exaggerating. Your solution seems correct, but for a beginner your solution seems a little bit confusing (Remember the questioner is a beginner), can you just organize a little bit your ideas, for example, the open set $U$ is the open set mentioned in the question before has edited? I hope you don't get angry with the downvotings and don't stop participating in stackexchange. I'm grateful with your answer, thank you! – user42912 Dec 13 '12 at 3:03
Instead of downvoting it, I suggest to instead comment "could you elaborate on...?" instead. I get no notification about you completely revamping your original question into a much better version, only the downvote. – gnometorule Dec 13 '12 at 16:41
To see (1), first convince yourself that (by the definition of a quotient map mapping saturated open sets to open sets) if you show that for any open $U$, $p^{-1}(p(U)$ is open again, the quotient map p is open, as in: maps open sets to open sets (this might take some thinking about it being true, but it is). Note that the antipodal map $a$, $a(x) = -x$, which induces the projection, is a homeomorphism (this should be clear). Then note that $p^{-1}(p(U)) = U \cup a(U)$, which is open. You now should have all the pieces in place to conclude - after thinking a bit. – gnometorule Dec 13 '12 at 18:10
yes, my mistake, I'm pretty new here, I'm sorry. Thank you for you answer, it helped a lot. – user42912 Dec 13 '12 at 21:52

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