Let a=(a_0, a_1, ..., a_n), b=(b_0, b_1$a=(a_0, a_1, ..., a_n )$, ..., b_n)$b=(b_0, b_1, ..., b_n )$ that belong to R^{n+1}${\mathbb R}^{n+1}$. Define polynomials f_a(t)=a_0+a_1t+ ... + a_nt^n$f_a (t)=a_0 +a_1 t+ ... + a_n t^n$ and f_b(t)=b_0+b_1t+ ... + b_nt^n$f_b (t)=b_0 +b_1 t+ ... + b_n t^n$ and let f_{ab}(t)=f_a(t)f_b'(t)-f_a'(t)f_b(t)=c_0+c_1t+ ... + c_{2n-2}t^{2n-2}$f_{ab}(t)=f_a (t)f_b '(t)-f_a '(t)f_b (t)=c_0 +c_1 t+ ... + c_{2n-2} t^{2n-2}$, where '$'$ denotes derivative. From the above setting we may define the (bilinear) map F$F$ :R^{2n+2} \to R^{2n-1}, F(a ${\mathbb R}^{2n+2} \to {\mathbb R}^{2n-1}$,b)=c $F(a,b)=c$.
QUESTION: Is F$F$ onto?
If so, is it known what the fiber F^{-1}(c)$F^{-1}(c)$ will look like?
