 Solution

For an affine point of $C$ with $x\ne 0$, the first coordinate of its image is $(y/x{)}^2$. Thus its image has trivial square class whenever that coordinate is nonzero. The identity and $T=(0,0)$ map to the identity. If a point maps to $T_1=(0,0)$, it has $y=0$ and $x^2+ax+b=0$, forcing $b_1=a^2-4b$ to be a rational square; in that case $q(T_1)=1$ too. Hence $\varphi (C(\mathbb{Q}))\subseteq \ker q$.

Conversely, let $(u,v)\in D(\mathbb{Q})$ have $q(u,v)=1$ and $u\ne 0$. Write $u=t^2$ with $t\in {\mathbb{Q}}^{\ast }$. The equation of $D$ givesSet $x=(u-a+v/t)/2$ and $y=tx$. Then $x$ solves $x^2+(a-u)x+b=0$ and is nonzero because $b\ne 0$. Hence $y^2=u{x}^2=x(x^2+ax+b)$, so $(x,y)\in C(\mathbb{Q})$. Also $x-b/x=v/t$, giving $\varphi (x,y)=(u,v)$.

The identity is an image of the identity. Finally, $T_1\in \ker q$ exactly when $b_1$ is a rational square; then either nonzero root $x=(-a\pm \sqrt{b_1})/2$ gives $(x,0)\in C(\mathbb{Q})$ mapping to $T_1$. This covers all exceptional points and proves