Let $a=(a_0, a_1, ..., a_n )$, $b=(b_0, b_1, ..., b_n )$ that belong to ${\mathbb R}^{n+1}$. Define polynomials $f_a (t)=a_0 +a_1 t+ ... + a_n t^n$ and $f_b (t)=b_0 +b_1 t+ ... + b_n t^n$ and let $f_{ab}(t)=f_a (t)f_b '(t)-f_a '(t)f_b (t)=c_0 +c_1 t+ ... + c_{2n-2} t^{2n-2}$, where $'$ denotes derivative. From the above setting we may define the (bilinear) map $F$ : ${\mathbb R}^{2n+2} \to {\mathbb R}^{2n-1}$, $F(a,b)=c$.

QUESTION: Is $F$ onto?

If so, is it known what the fiber $F^{-1}(c)$ will look like?

realsurjectivity, for which just a single differential is not sufficient. $\endgroup$ – Alex Degtyarev May 21 '16 at 10:02