Is the image of polynomial map of a box convex?

Question

Suppose I have $F = (F_1, F_2)$, where $F_i$ is a polynomial in two variables. Suppose further that the Jacobian of $F$ at $(0,0)$ is non-zero.
Let $B = (-1,1)^2$. Does it then follow that $F(B)$ is a convex set by any chance? (If not, are there conditions I can impose on $F$ such that this holds?)

Answer 1

No, but checking directly shows that any map of degree $1$ (i.e., affine map) preserves convexity, so any counterexample must have degree at least $2$.

Hint Consider the map $\Bbb C \to \Bbb C$ defined by $$\widetilde F : z \mapsto (z + 2)^2.$$ Under the usual identification $\Bbb R^2 \leftrightarrow \Bbb C$, $x + i y \leftrightarrow z$, this becomes $$F : (x, y) \mapsto (x^2 - y^2 + 4 x + 4, 2 x y + 4 y) .$$

Applying $F$ to the left corners of the box gives $F(-1, \pm 1) = (0, \pm 2)$. But the only solution to $\widetilde F(z) = 0$ is $z = -2$, so the only solution to $F(x, y) = (0, 0)$ is $(-2, 0)$, which is not in the closure of $F(B)$, so $F$ is not convex. On the other hand, since $F'(z) = 2 (z + 2)$, we have $(\operatorname{Jac} F)(x, y) = 4((x + 2)^2 + y^2)$, and so $\operatorname{Jac} F$ does not vanish at $(0, 0)$.