Exercise 11.8 of Cox's Primes of the forms $x^2+ny^2$ asks to prove that the map $$\Phi:C(m)\to\{\text{right cosets of $\Gamma_0(m)$ in $\Gamma$}\}, \sigma\mapsto(\sigma_0^{-1}\Gamma\sigma)\cap\Gamma$$ is bijective, where $\Gamma=\text{SL}(2,\mathbb Z)$ is the modular group, $\Gamma_0(m)$ is the Hecke congruence subgroup of level $m$, $\sigma_0= \begin{pmatrix}m&0\\0&1\end{pmatrix}$ and $$C(m)=\left\{ \begin{pmatrix}a&b\\0&d\end{pmatrix}:ad=m,a>0,d>b\geq0,\gcd(a,b,d)=1 \right\}.$$
I've been able to show that $\Phi(\sigma_0)=\Gamma_0(m)$, that $\Phi$ (if well-defined) is injective, and that $\Phi$ is well-defined (i.e., that $\Phi(\sigma)$ is actually a right coset) provided that $\Phi(\sigma)$ is nonempty. However,
Q1. How do you prove that $\Phi(\sigma)\neq\varnothing$?
Q2. How do you prove that $\Phi$ is surjective?
All my attempts have been unsuccessful. For Q1., I've tried to find an element $\gamma\in\Gamma$ such that $\sigma_0\gamma\sigma^{-1}\in\Gamma$, since that implies that $\sigma^{-1}\Gamma\sigma\cap\Gamma$ is nonempty. If we write $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $\sigma=\begin{pmatrix}a_0&b_0\\0&d_0\end{pmatrix}$, then
$$\sigma_0\gamma\sigma^{-1}= \begin{pmatrix}a d_0&a_0 b - a b_0\\c/a_0&(da_0-b_0 c)/(a_0 d_0)\end{pmatrix}$$
and $\sigma_0\gamma\sigma^{-1}\in\Gamma$ iff $c\in a_0\mathbb Z$ and $da_0-b_0 c\in a_0d_0\mathbb Z$ (and of course $ad-bc=1$). But I haven't been able to show that there exist $a,b,c,d\in\mathbb Z$ satisfying these conditions.