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Nevanlinna in his book Analytic functions seems to state the following (at the very end of Ch. X): For every compact Riemann surface $X$ of genus $g\geq 2$ there is a non-constant holomorphic map $f : X \to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

I explain in more detail. Nevanlinna proves the theorem of Picard that there is no non-constant map of the complex lane to a compact Riemann surface of genus $g \geq 2$. The proof begins with the sentence:

As Picard noticed, it is enough to prove this for a hyperelliptic curve.

And proceed to prove for hyperelliptic case.

The corresponding paper of Picard (Acta 11, 1887, p. 11-12) refers in the relevant place to a private letter of Hurwitz. It says:

Soit $f(x, y) = 0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consdquent $g>2$. A l'equation prcedente, le savant geomere associe une relation $$f_1(x,y_1)=0\quad\mbox{de genre}\quad g=2$$ jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y_1$ de $x$ sont tous compris parmi les points de ramification de a fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.

Can anyone explain what Picard says here, and why is this correct?

Nevanlinna in his book Analytic functions seems to state the following (at the very end of Ch. X): For every compact Riemann surface $X$ of genus $g\geq 2$ there is a non-constant holomorphic map $f : X \to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

I explain in more detail. Nevanlinna proves the theorem of Picard that there is no non-constant map of the complex plane to a compact Riemann surface of genus $g \geq 2$. The proof begins with the sentence:

As Picard noticed, it is enough to prove this for a hyperelliptic curve.

And proceeds to prove the theorem for hyperelliptic case.

The corresponding paper of Picard (Acta 11, 1887, p. 11-12) refers in the relevant place to a private letter of Hurwitz. It says:

Soit $f(x, y)=0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consequent $g>2$. A l'equation prcedente, le savant geomere associe une relation $$f_1(x,y_1)=0\quad\mbox{de genre}\quad g=2$$ jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y_1$ de $x$ sont tous compris parmi les points de ramification de la fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.

Can anyone explain what Picard says here, and why is this correct?

Nevanlinna in his book Analytic functions seems to state the following  (at the very end of Ch. X): For every compact Riemann surface $X$ of genus  $g\geq 2$ there is a non-constant holomorphic map $f:X\to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

I explain more detail. Nevanlinna proves there the theorem of Picard that there is no non-constant map of the complex lane to a compart Riemann surface of genus $g\geq2$. The proof begins with the sentence:

As Picard noticed, it is enough to prove this for a hyperelliptic curve.

And proceeds proving for hyperelliptic case.

The corresponding paper of Picard (Acta 11, 1887, p. 11-12) refers in the relevant place to a private letter of Hurwitz, and says:

Soit $f(x, y) = 0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consdquent $g>2$. A l'equation prcedente, le savant geomere associe une relation $$f_1(x,y_1)=0\quad\mbox{de genre}\quad g=2$$ jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y_1$ de $x$ sont tous compris parmi les points de ramification de a fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.

Can anyone explain what Picard says here, and why is this correct?

Nevanlinna in his book Analytic functions seems to state the following  (at the very end of Ch. X): For every compact Riemann surface $X$ of genus  $g\geq 2$ there is a non-constant holomorphic map $f : X \to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

I explain in more detail. Nevanlinna proves the theorem of Picard that there is no non-constant map of the complex lane to a compact Riemann surface of genus $g \geq 2$. The proof begins with the sentence:

As Picard noticed, it is enough to prove this for a hyperelliptic curve.

And proceed to prove for hyperelliptic case.

The corresponding paper of Picard (Acta 11, 1887, p. 11-12) refers in the relevant place to a private letter of Hurwitz. It says:

Soit $f(x, y) = 0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consdquent $g>2$. A l'equation prcedente, le savant geomere associe une relation $$f_1(x,y_1)=0\quad\mbox{de genre}\quad g=2$$ jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y_1$ de $x$ sont tous compris parmi les points de ramification de a fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.

Can anyone explain what Picard says here, and why is this correct?

Nevanlinna in his book Analytic functions states the following (at the very end of Ch. X): For every compact Riemann surface $X$ of genus $g\geq 2$ there is a non-constant holomorphic map $f:X\to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

Nevanlinna in his book Analytic functions seems to state the following (at the very end of Ch. X): For every compact Riemann surface $X$ of genus $g\geq 2$ there is a non-constant holomorphic map $f:X\to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

I explain more detail. Nevanlinna proves there the theorem of Picard that there is no non-constant map of the complex lane to a compart Riemann surface of genus $g\geq2$. The proof begins with the sentence:

As Picard noticed, it is enough to prove this for a hyperelliptic curve.

And proceeds proving for hyperelliptic case.

The corresponding paper of Picard (Acta 11, 1887, p. 11-12) refers in the relevant place to a private letter of Hurwitz, and says:

Soit $f(x, y) = 0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consdquent $g>2$. A l'equation prcedente, le savant geomere associe une relation $$f_1(x,y_1)=0\quad\mbox{de genre}\quad g=2$$ jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y_1$ de $x$ sont tous compris parmi les points de ramification de a fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.

Can anyone explain what Picard says here, and why is this correct?