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Is it true that for any non-orientable Riemannian surface of genus 2 there exists Conformal mapping of degree two to a projective plane?

Also, is the following argument works? Given any Riemann surface $M$ of genus two with a free involition $\sigma$ we can find a meromorphic $f$ function with the only pole of second order (existence of Weierstrass point). Then we consider the function $f +\sigma^*f$, it is a conformal mapping to a Riemann sphere of degree 4 and it descends to a factor $M/\sigma$. As a result we have a conformal mapping of order 2 of any non-orientable to a Riemann sphere. I have some suspicions about the result, but do not seem to find an error.

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What's your definition of a conformal mapping, especially in view of the fact that your maps are branched covers? If it means a map preserving nonoriented angles away from branch-points, then the answer is negative: Such map does not exist (generically) once $\chi(M)<-2$. – studiosus Oct 19 '13 at 21:41

Judging from your proposed solution, when you refer to a "non-orientable Riemann surface $M$ of genus 2" what you have in mind is an orientable genus 2 surface equipped with a fixed-point-free involution $\tau$ which reverses orientation. If this is the case, one can argue as follows. Let $J$ be the hyperelliptic involution of $M$. It is known that $\tau$ and $J$ necessarily commute. Therefore $\tau$ descends to the quotient of $M$ by $J$, i.e., the Riemann sphere. The quotient of the Riemann sphere by $\tau$ gives the real projective plane. All maps involved are locally holomorphic or anti-holomorphic, giving the result. A more detailed discussion may be found here.

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