Show a map from $\mathbb{P}^2$ to $\mathbb{P}^2$ is injective

Question

I have a map $F: \mathbb{P}^2 \to \mathbb{P}^2$ defined by $F(x_0 : x_1 : x_2) = (x_0^2 + x_2^2 : x_0 x_1 + x_1 x_2 : x_0 x_2 + x_1^2)$ and I need to show it is well defined and smooth. I already proved this, but now I need to show that the restriction of $F$ to the chart $U_0 = \{(x_0 : x_1 : x_2) \in \mathbb{P}^2 | x_0 \neq 0\}$ is injective. I have tried to put $F(x_0 : x_1 : x_2) = F(y_0 : y_1 : y_2)$ and deduce that $(x_0 : x_1 : x_2) = (y_0 : y_1 : y_2)$ (as points of $\mathbb{P}^2$), but couldn't conclude. Any suggestion?

Answer 1

If a point $p\in U_0\implies F(p)\in\tilde {U_0}$, so the local representative of the map is $\hat F=\tilde \psi_0\circ F\circ \psi_0$.
If we take a point $(u,v)\in\psi_0(U_0)$, we have $$\hat F(u,v)=\tilde \psi_0\circ F(1:u:v)=\tilde \psi_0(1+v^2:u+uv:v+u^2)= \bigg(\dfrac{u(v+1)}{1+v^2},\dfrac{v+u^2}{1+v^2}\bigg)\implies$$$$ \hat F \in\mathcal C^{\infty}(\psi_0(U_0)).$$ The map $F_{*p}:T_p\mathbb P^2\to T_{F(p)}\mathbb P^2$ is locally represented by $$J_{\hat F}\vert_p=\begin{pmatrix}\nabla \hat F_1(u,v)\\\nabla\hat F_2(u,v) \end{pmatrix}\Bigg\vert_p$$ but if we evaluate $J_{\hat F}$ in $(0,-1)\implies \text {rk}(J_{\hat F})<2\implies F$ is not an immersion.
Take now two points $(1:1:2)$, then $F(1:1:2)=(5:3:3)=F(2:1:1)=(5:3:3)$ but since $\nexists\lambda\in\mathbb R$ s.t. $(2:1:1)=\lambda(1:1:2)\implies F$ can't be an injective immersion.