This is problem $2.11$ page $133$ of Kenji Ueno's book.
Consider an irreducible hypersurface $V(F)$ where we assume that the homogeneous polynomial of degree $d$ satisfies the conditions $F(0,x_{1},..,x_{n}) \neq 0$ and $F(1,0,...0) \neq 0$.
Let $P=(1:0:...:0)$. For a point $Q \in V(F)$ let $R(Q)$ be the intersection of the line $PQ$ with the hyperplane $H_{\infty}: x_{0}=0$. Let $\phi_{P}: Q \in V(F) \rightarrow R(Q) \in H_{\infty}$.
Prove that for a point $R \in H_{\infty}$ there is a point $Q \in V(F)$ such that $R(Q)=R$.
I'm not sure how to write this. I tried looking at an example, say we are in $\mathbb{P}^{2}$ and $F(x,y,z)=xz+y^{2}$ then $\phi_{P}$ is the map that sends $(x:y:z)$ to $(0:y:z)$.
So suppose we have a point $(0:b:c) \in H_{\infty}$, we want to find a point $(u:v:w) \in V(xz+y^{2})$ such that $\phi_{P}(u:v:w)=(0:b:c)$. Now take $b=1,c=3$.
This implies there is some nonzero $\lambda$ such that $v=\lambda b$ and $w= \lambda c$. Since $b=1$ and $c=3$ then $v=\lambda$ and $w=3\lambda$ so we need to solve $uw+v^{2}=0$. Plugging these values yields:
$u(3\lambda)+\lambda^{2}=0$
So take $u=(-1/3)\lambda$. Thus setting $\lambda=-3$ gives $u=1$, $v=-3$, $w=-9$.
Then $(u:v:w) \in V(F)$ and the map sends this point to $(0:-3:-9)=(0:1:3)$ as desired.
How can we write this in a general way?