Let $h,g$ be polynomials, can function $f(x) = (h(x), g(x))$ be injective? Let $h,g$ be not injective functions, can function $f:\mathbb{R}\rightarrow \mathbb{R}^2$ such that $f(x) = (h(x), g(x))$ be injective?
I know that, if I pick polynomials for $h$ and $g$, then may be not injective, if picked carefully. For example, I can check zeros of the polynomials, whether they collide or not, but I am really not sure whether there exist other, nontrivial counterexample to disprove injectivity of $f$.

$f$ can be injective. For instance, $g(x) = x^2, g(x) = (x + 1)^2$. Then $f(x) = (x^2, (x+1)^2)$ is injective, since $g(a) = g(b)$ and $a \neq b$ means $a = -b$, but $h(a) \neq h(-a)$. Thus, if $a$ and $b$ are such that the first component of $f(a)$ is equal to the first component of $f(b)$, then the second components are different.
As for a non-trivial, non-injective $f$, one may take $g(x) = x^2$ and $h(x) = x^3  - x$. Then $f(-1) = f(1) = (1, 0)$.
