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EDIT: The well known Jordan curve theorem says: let $C\subset S^2$ be a closed simple curve on the 2-sphere. Then its complement $S^2\backslash C$ consists of two connected components, both homeomorphic to discs (in fact it is known that the closure of each component is homeomorphic to the closed disk by Jordan-Schoenflies theorem).

Is there a version of the Jordan theorem for closed simple curves in real projective plane $\mathbb{R}\mathbb{P}^2$? (The curve might be assumed to be smoothly imbedded.)

A reference would be helpful.

ADDED: Given the comment by HenrikRüping below, I realized that for my purposes it suffices to assume that the homology class of $C$ vanishes in $H_1(\mathbb{R}\mathbb{P}^2,\mathbb{Z}/2\mathbb{Z})$.

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    $\begingroup$ of course if we look at a closed simple curve representing the generator of $\pi_1$, the complement should have only one connected component homeomorphic to a disc. $\endgroup$ Dec 15, 2019 at 18:40
  • $\begingroup$ @HenrikRüping: Many thanks. Corrected. $\endgroup$
    – asv
    Dec 15, 2019 at 18:43

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If C is null-homologous, then the complement of C has two components: a disk and a Möbius strip (as one sees since the preimage of C in the 2-sphere is 2 disjoint Jordan curves). If C is not null-homologous, then the complement of C is a single disk.

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  • $\begingroup$ Is there a reference for this? (I think I would be able to prove it, but I prefer to just cite it) $\endgroup$
    – Saúl RM
    Nov 4, 2022 at 21:38
  • $\begingroup$ You can probably make an easy proof just with the elementary theory of coverings and the Euler characteristic of the complement. $\endgroup$ Feb 21, 2023 at 20:25
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Well, if you take the double cover, under your assumptions the lift is two simple closed curves in $S^2,$ the complement of which will be two disks and an annulus, so the original curve bounds a disk on one side.

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