Number of cusps of an modular curve $X_0(N)$

Question

Let $X_0(N) = \Gamma_0(N) / (\mathbb{H} \cup \mathbb{P}^1(\mathbb{Q}))$. A lecture note (p. 2) lists the following (very easy!) formula for the corresponding cusps:

$$\nu_\infty = \sum_{d\mid N} \varphi(\gcd(d,N/d)).$$

Unfortunately, there is no source or derivation given for this formula. I know what a cusp is and that the set of cusps is finite for all subcongruence groups of $\operatorname{SL}_2(\mathbb{Z})$. However, I do not grasp why this formula for $\nu_\infty$ is correct.

Answer 1

See Section 3.8 of Diamond and Shurman's "A first course in modular forms", page 103.

Alternatively, see Prop. 1.43 in page 24 of Shimura's "Introduction to the arithmetic theory of automorphic functions".

Answer 2

This is proposition 2.2 in Yuri Manin's paper 'Parabolic points and zeta functions of modular curves'. He proves a bit more namely that the map $$ \frac{u}{v\delta}\mapsto [\delta,uv\mod \gcd(\delta,N\delta^{-1})], \quad \infty\mapsto[N,1 ] $$ is a bijection, the domain being $\Gamma_0(N)\backslash P^1(\mathbb{Q})$ the cusps of $\Gamma_0(N)$ and the codomain the pairs $[\delta,x\mod\gcd(\delta,N\delta^{-1})]$, where $\delta$ is a divisor of $N$ and $x$ is invertible modulo $\gcd(\delta,N\delta^{-1}]$. $\delta$ is chosen so that $\gcd(u,v\delta)=\gcd(v,N\delta^{-1})=1$. The second set in this bijection obviously has the right cardinality.