Is the 1 dimensional projective line homeomorphic to the circle? If so, the circle is homeomorphic to itself with antipodal points identified (very unintuitive). I am missing something?
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They are homeomorphic. Why do you think you are missing something? – Michael Albanese Mar 13 '17 at 23:41
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2My intuition would say that like the sphere, if you identify antipodal points you should change the structure, not end up with what you started with. – PossumP Mar 13 '17 at 23:51
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4@user54301 Have you ever taken a rubber band and doubled it to form a smaller loop? It's pretty easy for a rubber band, but not so obvious for a rubber sphere :). – Erick Wong Mar 13 '17 at 23:58
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Yes, the real $1$-dimensional projective line $\mathbb{RP}^1$ is homeomorphic (in fact diffeomorphic) to the circle $S^1$. It can be thought of as the circle $S^1$ with antipodal points identified, which reflects the fact that the circle is its own double cover: the double cover map is given explicitly by
$$S^1 \ni z \mapsto z^2 \in S^1$$
thinking of $S^1$ as the unit complex numbers. In other words, quotienting by antipodes is the same as quotienting by $180^{\circ}$ rotation; note that this is very much false for the 2-sphere.
Qiaochu Yuan
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I see each point, individually. Should I be thinking of the nature projective planes as closely related to covers? Even for non-spheres? – PossumP Mar 14 '17 at 00:28
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Here is a somewhat lower-brow 'proof by picture' of how such a process of gluing might go. To save you from having to read my poor handwriting, the steps are:
1) Embed $S^1$ into the $(x,y)$-plane in $\mathbb{R}^3$.
2) Pick an antipodal pair and glue (together) to the $z$-axis.
3) Pick another pair of non-glued antipodal points, and snip the two loops at this pair.
4) Glue the antipodal pairs of line segments (together) along the $z$-axis. You'll have but one pair left unglued.
5) Glue final pair.
Pete Caradonna
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