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Why in proof of proposition 6 of http://arxiv.org/abs/0909.1665, they claim that if a embedded surfaces $\Sigma^2 \subset (M^3,g)$ is homeomorphic to $\mathbb{RP}^2$, where $M$ is compact manifold, then, by uniformization theorem, there exist diffeomorphism $\phi: \mathbb{RP}^2 \rightarrow \Sigma$ such that $\phi^{*}g$ is conformal to standard metric on $\mathbb{RP}^2$?

Can someone help me?

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    Because there is a unique conformal structure on the projective plane. That's a consequence of the uniformization theorem.2012-12-20

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