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The space $P^1$ and the covering map $p:S^1\rightarrow P^1$ are familiar ones. What are they?

$S^1=\{z \in \mathbb{C}\mid |z|=1\}$ and $P^1$ is the quotient space where we take $S^1$ and identify $z \in S^1$ with $-z \in S^1$. In a previous assignment, we showed that the map $q:S^1\rightarrow S^1$ by $q(z)=z^2$ is a covering map of $S^1$. I see that $q(z)=z^2=(-z)^2=q(-z)$ so $z$ and $-z$ get mapped to the same point. This is supposed to be helpful for showing that then $P^1$ is homeomorphic to $S^1$. To me, $P^1$ looks like just for example the upper half of the circle $S^1$. I can imagine this being continuously mapped onto $S^1$ so I see how they could be homeomorphic but I am not sure how to use $q$ being a covering map that maps antipodal points to the same image to conclude that $P^1$ is $S^1$. I am a little lost =/

2 Answers2

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This will help more with your intuition rather than being a rigorous answer, but you can think of $\Bbb R P^1$ as the upper half of the circle, but remember that the two endpoints of this semicircle get glued together, giving you something homeomorphic to a circle.

Dylan Yott
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Consider the map $z \mapsto z^2$ from $S^1 \to S^1$. By the universal property of the quotient topology on $P^1$, this drops to a continuous bijection from $P^1 \to S^1$. Now, you just need to check that this is also open.

ronno
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  • I see that the map which you said is a continuous bijection is an open map. So I get that $P^1$ is homeomorphic to $S^1$. I am confused about when the question mentions "the covering map $p:S^1\rightarrow P^1$." There is some canonical covering map? I know the quotient map $S^1 \rightarrow P^1$. Quotient maps going from the set $S^1$ to the set of equivalence classes $P^1$ is the covering map the question is mentioning? Thanks!! –  Dec 05 '13 at 04:34
  • Yes, the quotient map is a $2$-sheeted covering map, a fact you can easily check. – ronno Dec 05 '13 at 04:54