Note that, since $H^{\circ}$ is defined over $\mathbb{R}$, we have $\gamma(\mathbb{R}H^{\circ})\subset{\mathbb{R}}{\mathbb{P}}^{n}$. Note that $\gamma$ extends to a map $H\to{\mathbb{C}}{\mathbb{P}}^{n}$ which is an isomorphism, since $\deg\gamma=1$. ∎