We first show that the system of equations implies $z_{1}z_{2}+z_{3}^{k}=0$, so that $\overline{\mathcal{S}}$ does project to $\mathcal{S}$. To see this, we notice the first three equations imply

(7.9)

$$u_{3}^{2}(z_{1}z_{2}+z_{3}^{k})=0.$$

If $u_{3}\neq 0$, then we get $z_{1}z_{2}+z_{3}^{k}=0$. If $u_{3}=0$, then by the third equation we get that either $u_{1}\neq 0,u_{2}=0$ or $u_{1}=0,u_{2}\neq 0$. In the first case using the remaining equations we get $z_{3}=z_{2}=0$. In the second case we get $z_{3}=z_{1}=0$. In both cases the equation $z_{1}z_{2}+z_{3}^{k}=0$ is indeed satisfied.

Now we study singularities of $\overline{\mathcal{S}}$. In the affine chart $\{u_{1}\neq 0\}$, we get

(7.10)

$$\begin{cases}v_{2}+v_{3}^{2}=0;\\ z_{2}+z_{3}^{a_{2}}v_{3}=0,\end{cases}$$

so we reduce the defining equations to a single equation in the $z_{1},z_{3},v_{3}$ variable given by

(7.11)

$$z_{3}^{a_{1}}=z_{1}v_{3}.$$

This has exactly one $A_{a_{1}-1}$ singularity at $\{z_{1}=z_{3}=v_{3}=0\}$. Similarly, on the affine chart $\{u_{2}\neq 0\}$ we reduce the equations to

(7.12)

$$z_{3}^{a_{2}}=z_{2}v_{3}.$$

This has exactly one $A_{a_{2}-1}$ singularity at $\{z_{2}=z_{3}=v_{3}=0\}$. On the affine chart $\{u_{3}\neq 0\}$, we reduce the equations to

(7.13)

$$v_{1}v_{2}+1=0.$$

which is smooth.

It is then easy to verify that the projection map $\overline{\mathcal{S}}\rightarrow\mathcal{S}$ is an isomorphism outside the point $\{z_{1}=z_{2}=z_{3}=0\}$, and if $z_{1}=z_{2}=z_{3}=0$, we get the equation

(7.14)

$$u_{1}u_{2}+u_{3}^{2}=0,$$

which gives a conic in $\mathbb{C}\mathbb{P}^{2}$. ∎