I am reading Shurman's A First Course in Modular Forms, and I got stuck in the argument to count cusps of $\Gamma_0(N)$ in page 103, here is the full argument for reference:
To count the cusps of $Γ_0 (N)$ recall from Proposition 3.8.3 that for this group, vectors $\begin{bmatrix}a \\ c\end{bmatrix}$ and $\begin{bmatrix}a' \\ c'\end{bmatrix}$ with $\gcd(a,c) = \gcd(a',c') = 1$ represent the same cusp when $\begin{bmatrix}ya' \\ c'\end{bmatrix} \equiv \begin{bmatrix}a +jc \\ c\end{bmatrix}\bmod{N}$ for some $j$ and $y$ with $\gcd(y,N)=1$. The bottom condition, $c'\equiv yc\mod{N}$ for some such $y$, is equivalent to $\gcd(c',N) = \gcd(c,N)$, in wich case letting $d = \gcd(c,N)$ and letting $y_0 ∈ ℤ$ satisfy $y_0 \equiv c'c^{-1} \bmod{N/d}$ makes the condition equivalent to $y \equiv y_0 + iN/d \bmod{N}$ for some $i$ (confirming the calculatons in the paragraph is Exercise 3.8.4.). For any divisor $d$ of $N$, pick one value $c$ modulo $N$ such that $\gcd(c,N) = d$. Then any cusp of $Γ_0(N)$ represented by some Vector $\begin{bmatrix}a' \\ c'\end{bmatrix}$ with $\gcd(c',N) = d$ is also represented by $\begin{bmatrix}a \\ c\end{bmatrix}$ whenever $(y_0 + iN/d)a' \equiv a+jc \bmod{N}$ for some $i$ and $j$, or $a \equiv y_0a' \bmod \gcd(c,N,a'N/d)$, or $a \equiv y_0 a' \bmod{\gcd(d,N/d)}$. Also, $a$ is relatively prime to $d$ since $\gcd(a,d) \mid \gcd(a,c) = 1$, so $a$ is relatively prime to $\gcd(d,N/d)$. Thus for each divisor $d$ of $N$ there are $\phi(\gcd(d,N/d))$ cusps, and the number of cusps of $\Gamma_0 (N)$ is $$\epsilon_{\infty}(\Gamma_0(N))=\sum_{d \mid N} \phi(\gcd(d,N/d)).$$
I have trouble in the sentence:
"makes the condition equivalent to $y\equiv y_0+iN/d \bmod{N}$, for some $i$".
It seems that what we should do here is, suppose vectors $\begin{bmatrix}a \\ c\end{bmatrix}$ and $\begin{bmatrix}a' \\ c'\end{bmatrix}$ with $\gcd(a,c) = \gcd(a',c') = 1$ represent the same cusp, then
(a)$\begin{bmatrix}ya' \\ c'\end{bmatrix} \equiv \begin{bmatrix}a +jc \\ c\end{bmatrix}\bmod{N}$ for some $j$ and $y$ with $\gcd(y,N)=1$.
is equivalent to
(b)$y \equiv y_0 + iN/d \bmod{N}$ for some $i$, where $y_0 ∈ ℤ$ satisfies $y_0 \equiv c'c^{-1} \bmod{N/d}$, and $d=\gcd(c,N)$.
and I got stuck in the proving this. Any help?