show that if the fiber $\pi_2^{-1}(\pp)$ of $\pp$ is of dimension 1, then $$\corank(M_{\nu}(\phi)(\pp))=HP_{\pi_2^{-1}(\pp)}(\nu)\geq \nu+1.$$ So, assume that $\pi_2^{-1}(\pp)$ is 1-dimensional and set $\delta:=\deg(\pi_2^{-1}(\pp))$ so that $$HP_{\pi_2^{-1}(\pp)}(\nu)=\delta\nu+1-\binom{\delta-1}{2}+N$$ where $\binom{\delta-1}{2}$ is the arithmetic genus of the unmixed dimension one part of the fiber $\pi_2^{-1}(\pp)$ and $N\geq 0$ is the degree of its remaining finite part. Observe that $\delta\leq d$. Now, a straightforward computation yields $$HP_{\pi_2^{-1}(\pp)}(\nu)=\frac{\delta(2\nu+3-\delta)}{2}+N.$$ From here, we see that $HP_{\pi_2^{-1}(\pp)}(\nu)$ is an increasing function of $\delta$ on the interval $[-\infty;\nu+\frac{3}{2}]$ which takes value ${\nu+1+N}$ for $\delta=1$. Observe moreover that $\nu+\frac{3}{2}\geq \delta$ as soon as $\nu \geq d-1$. And if $\nu=d-2$, then $HP_{\pi_2^{-1}(\pp)}(d-2)$ takes the same value for $\delta=d-1$ and $\delta=d$.

An interesting consequence of the above proposition is that for all integer $0\leq n \leq 2d-2$ $$\corank(M_{2d-2}(\phi)(\pp))=n \Leftrightarrow \pi_2^{-1}(\pp) \textrm{ is finite of degree } n.$$ Therefore, as soon as $d\geq 2$ we have $$\corank(M_{2d-2}(\phi)(\pp))=1 \Leftrightarrow \pp \textrm{ has a unique preimage}.$$ Observe in addition that once we know that a point $\pp$ has a unique preimage, this preimage can be computed from the kernel of the transpose of $M_{2d-2}(\phi)(\pp)$. Indeed, this kernel is generated by a single vector. But it is clear by the definition of $M_{2d-2}(\phi)$ that the 