Bijection between the set of classes of positive definite quadratic forms and the set of classes of quadratic numbers in the upper half plane Let $\Gamma = SL_2(\mathbb{Z})$.
Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$.
Let $f(x, y) = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $\Gamma$.
We write $f^\alpha(x, y) = f(px + qy, rx + sy)$.
Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $\Gamma$ acts on $\mathfrak{F}$.
Let $f, g \in \mathfrak{F}$.
If $f$ and $g$ belong to the same $\Gamma$-orbit, we say $f$ and $g$ are equivalent.
Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
We say $D = b^2 - 4ac$ is the discriminant of $f$.
It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $f$ is primitive.
If $D < 0$ and $a > 0$, we say $f$ is positive definite.
We denote the set of positive definite primitive binary quadratic forms of discriminant $D$ by $\mathfrak{F}^+_0(D)$.
By this question, $\mathfrak{F}^+_0(D)$ is $\Gamma$-invariant.
We denote the set of $\Gamma$-orbits on $\mathfrak{F}^+_0(D)$ by $\mathfrak{F}^+_0(D)/\Gamma$.
Let $\mathcal{H} = \{z \in \mathbb{C}; Im(z) > 0\}$ be the upper half complex plane.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $\Gamma$.
Let $z \in \mathcal{H}$.
We write $$\alpha z = \frac{pz + q}{rz + s}$$
It is easy to see that $\alpha z \in \mathcal{H}$ and $\Gamma$ acts on $\mathcal{H}$ from left.
Let $\alpha \in \mathbb{C}$ be an algebraic number.
If the minimal plynomial of $\alpha$ over $\mathbb{Q}$ has degree $2$, we say $\alpha$ is a quadratic number.
Let $\alpha$ be a quadratic number.
There exists the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$.
$D = b^2 - 4ac$ is called the discriminant of $\alpha$.
Let $\alpha \in \mathcal{H}$ be a quadratic number.
There exists the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$.
Let $D = b^2 - 4ac$.
Clearly $D < 0$ and $D$ is not a square integer.
It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
Conversly suppose $D$ is a negative non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
Then there exists a quadratic number $\alpha \in \mathcal{H}$ whose discriminant is $D$.
We denote by $\mathcal{H}(D)$ the set of quadratic numbers of discriminant $D$ in $\mathcal{H}$.
It is easy to see that $\mathcal{H}(D)$ is $\Gamma$-invariant.
Hence $\Gamma$ acts on $\mathcal{H}(D)$ from left.
We denote the set of $\Gamma$-orbits on $\mathcal{H}(D)$ by $\mathcal{H}(D)/\Gamma$.
Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}^+_0(D)$.
We denote $\phi(f) = (-b + \sqrt{D})/2a$, where $\sqrt{D} = i\sqrt{|D|}$.
It is clear that $\phi(f) \in \mathcal{H}(D)$.
Hence we get a map $\phi\colon \mathfrak{F}^+_0(D) \rightarrow \mathcal{H}(D)$.
My question
Is the following proposition true?
If yes, how do we prove it?
Proposition
Let $D$ be a negative non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
Then the following assertions hold. 
(1) $\phi\colon \mathfrak{F}^+_0(D) \rightarrow \mathcal{H}(D)$ is a bijection.
(2) $\phi(f^\sigma) = \sigma^{-1}\phi(f)$ for $f \in \mathfrak{F}^+_0(D), \sigma \in \Gamma$.
Corollary
$\phi$ induces a bijection $\mathfrak{F}^+_0(D)/\Gamma \rightarrow \mathcal{H}(D)/\Gamma$.
 A: Proof of (1)
We define a map $\psi\colon \mathcal{H}(D) \rightarrow \mathfrak{F}^+_0(D)$ as follows.
Let $\theta \in \mathcal{H}(D)$.
$\theta$ is a root of the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$.
$D = b^2 - 4ac$.
We define $\psi(\theta) = ax^2 + bxy + cy^2$.
Clearly $\psi$ is the inverse map of $\phi$.
Proof of (2)
Let $f = ax^2 + bxy + cy^2$.
Let $\sigma = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right) \in \Gamma$.
Let $f^\sigma = kx^2 + lxy + my^2$.
Let $\theta = \phi(f)$.
Let $\gamma = \sigma^{-1}\theta$.
Then $\theta = \sigma \gamma$.
$a\theta^2 + b\theta + c = 0$.
Hence $a(p\gamma + q)^2 + b(p\gamma + q)(r\gamma + s) + c(r\gamma + s)^2 = 0$.
The left hand side of this equation is $f(p\gamma + q, r\gamma + s) = k\gamma^2 + l\gamma + m$.
Since $f^\sigma$ is positive definite by this question, $k \gt 0$.
Since gcd$(k, l, m)$ = 1 by the same question, $\psi(\gamma) = f^\sigma$.
Hence $\phi(f^\sigma) = \gamma = \sigma^{-1}\theta$.
This proves (2).
