Exercise 10.2

Proof. Note that if one representative $(x_0,\dots ,x_n)$ of a projective point satisfies $x_0=0$, then it is the same for all representatives of this point, so we can define

where we write for simplicity $[x_0,\dots ,x_n]$ for $[(x_0,\dots ,x_n)]$.

Consider

Then $\psi$ is well-defined. Indeed, if $(0,x_1,\dots ,x_n)\sim (0,y_1,\dots ,y_n)$, then there is some $\lambda \in F^{\text{∗}}$ such that $(0,y_1,\dots ,y_n)=\lambda (0,x_1,\dots ,x_n)$, thus $(y_1,\dots ,y_n)=\lambda (x_1,\dots ,x_n)$, and $[x_1,\dots ,x_n]=[y_1,\dots ,y_n]$.

If $\psi ([0,x_1,\dots ,x_n])=\psi ([0,y_1,\dots ,y_n])$, then $[(x_1,\dots ,x_n)]=[(y_1,\dots ,y_n)]$, so there is some $\lambda \in F^{\text{∗}}$ such that $y_i=\lambda x_i,i=1,\dots ,n$. Since $0=\lambda 0$, $(0,y_1,\dots ,y_n)\sim (0,x_1,\dots ,x_n)$, therefore $[0,x_1,\dots ,x_n]=[0,y_1,\dots ,y_n]$, so $\psi$ is injective.

Moreover if $[x_1,\dots ,x_n]$ is any projective point of $P_{n-1}(F)$, then $[x_1,\dots ,x_n]=\psi ([0,x_1,\dots ,x_n])$ so $\psi$ is surjective.

To conclude, $\psi$ is a bijection. □