Start from a map $\mathbb{CP}^1 \ni (u:v) \mapsto (x:y:z) \in \mathbb{CP}^2$ given by $$ x = au^2, \quad y = av^2, \quad z = uv,$$ where $a \in \mathbb{R}$ is a parameter. The image of the map is the conic $$ xy - a^2z^2=0,$$ which is real under $(x:y:z) \mapsto (\overline{y}:\overline{x}:\overline{z}).$
To make the map equivariant, we choose to act on $\mathbb{CP}^1$ with the involution $$ \Omega:(u:v) \mapsto (\overline{v}:\overline{u}).$$
The statement I want to prove or disprove is that we obtain a map from the disk to $\mathbb{CP}^2$ (and this is OK) with boundary on $\mathbb{RP}^2.$
I'm not fully convinced that the boundary is on $\mathbb{RP}^2:$ since the equation reduces by one the number of complex variables, and then reality is imposed, I would expect the boundary to be mapped to $\mathbb{RP}^1.$
In other terms, calling $C=\left\{ (x:y:z) \in \mathbb{CP}^2 \left| xy = a^2z^2 \right. \right\}$ the boundary should go to $L,$ where $L=\left\{ x \in C \left| \Omega(x)=x \right.\right\}$ is topologically $\mathbb{RP}^1.$