The modular curve $X_0(11)$ has a quotient isomorphic to Riemann sphere Consider the congruence subgroup
$$\Gamma_0(11)=\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in SL(2,\mathbb{Z}) \right\}$$
and the associated modular curve $X_0(11).$ I can prove that the Fricke involution $\omega_{11}:H\to H$ (where $H$ is the Poincaré half-plane)
$$\tau\mapsto-\frac{1}{11\tau}$$
induces an involution $\omega_{11}$on the modular curve $X_0(11).$ I can consider the quotient $X_0^+(11)=X_0(11)/\langle\omega_{11}\rangle,$ which is also a compact Riemann surface. The projection $\pi:X_0(11)\to X_0^+(11)$ has degree $2.$ I want to prove that $X_0^+(11)$ has genus $0,$ and hence is isomorphic to the Riemann sphere.
I want to prove this fact using the Hurwitz formula. The only thing I miss is how to calculate the points with multiplicity. How can I do that?
 A: We know already that $X_0(11)$ has genus $1$. Now in general suppose that $X$ is a curve (Riemann surface etc) of genus $1$, and that there is a degree $2$ map $X \to X'$ where $X'$ has genus $0$.
By Riemann-Hurwitz you should have
$$2g - 2 = 2(2g' - 2) + \sum_p(e_p - 1).$$
So there should be 4 ramification points (since $0 = -4 + \sum(e_p -1)$).

Aside: One can think about this geometrically - let $E$ be an elliptic curve with a degree $2$ mapping to a copy of $\mathbb{P}^1$, then choosing this $\mathbb{P}^1$ as the $x$-coordinate we see this double cover is ramified above the 2-torsion points (this is essentially what we do when we put $E$ in Weierstrass form).

Now to the original question, let $g = g(X_0(11))$ and $g^+ = g(X_0^+(11))$. You know that $2g -2 = 0$, and you know there is a ramification point (clearly $\sqrt{-1/11}$ works, as noted in the comments). Therefore the sum $ \sum_p(e_p -1 ) \geq 1$.
In particular we then must have $2(2g^+ - 2) < 0$, which can only occur when $g^+ <1$. However $g^+$ is a nonnegative integer, so it must be zero, and there must be $4$ ramification points.
