Let $F: \mathbb{R}^2 \to \mathbb{R}^2$, $(x,y) \mapsto (f(x,y),g(x,y))$, where $f(x,y),g(x,y) \in \mathbb{R}[x,y]$, each is of $(1,1)$-degree at least one, and $f(x,y),g(x,y)$ are algebraically independent over $\mathbb{R}$ (their Jacobian $\in \mathbb{R}[x,y]-\{0\}$).
Assume that for every $a,b,\lambda \in \mathbb{R}$, if $F(a,\lambda)=F(b,\lambda)$ then $a=b$, namely, $F$ is injective on horizontal lines $y=\lambda \in \mathbb{R}$.
(1) It would be nice to have an example of such $F$ which is non-injective. (I guess there exists such $F$ which is non-injective?).
A second question: Assume that for every $a,b,\lambda \in \mathbb{R}$, if $F(a,\lambda a)=F(b,\lambda b)$ then $a=b$, namely, $F$ is injective on lines through the origin.
(2) It would be nice to have an example of such $F$ which is non-injective. (I guess there exists such $F$ which is non-injective?).
A third question: $F$ which satisfies both the first and second conditions, still does not have to be injective?
Thank you very much!
Let $F$ be given by $$F(x,y)=(x^3+y^3,2x+y)$$
Then $F$ is injective on every horizontal line, and on every line through the origin, but $F$ is not injective, since for example, the equation $$F(x,y)=\bigl({\small{\frac{1}{4}}},1\bigr)$$ has more than one solution.