Let $I':=(g_1,g_2,g_3)$. We may assume for simplicity that $\pp=(1:0:0:0)$ (i.e. $f=f_0$ and $f_i=hg_i$ for $i=1,2,3$), and we then have an exact sequence $$0 \rightarrow Z_1'(d) \rightarrow Z_1(d) \rightarrow S \rightarrow {\mathcal S}_I\otimes \kappa(\pp) \rightarrow 0,$$ with $Z_1'$ the syzygy module of $(f_1,f_2,f_3)$, the first map sending $(b_1,b_2,b_3)$ to $(0,b_1,b_2,b_3)$ and the second $(a_0,a_1,a_2,a_3)$ to $a_0$.

It follows that $\operatorname{reg}( {\mathcal S}_I\otimes \kappa(\pp))\leq \max\{ \operatorname{reg}(S),\operatorname{reg}(Z_1)-d-1,\operatorname{reg}(Z'_1)-d-2\}$ and $a^0 ( {\mathcal S}_I\otimes \kappa(\pp))\leq a^2(Z'_1)-d\leq \operatorname{reg}(Z'_1)-d-2$ as $H^0_{\mathfrak m}(S)=H^1_{\mathfrak m}(Z_1)=0$.

By Lemma \[1fiber\], $h=\gcd (f_1,f_2,f_3)$ is of degree $\delta$. Set $f_i=hf_i'$ for all $i=1,2,3$. The syzygies of $(f_1,f_2,f_3)$ and $I':=(f_1',f_2',f_3')$ are equal, up to a degree shift by $\delta$, and the exact sequence $0\to Z'_1 (\delta )\to S(-d+\delta )^3 \to I'\to 0$ shows that : $$\operatorname{reg}(Z'_1)=\operatorname{reg}(I')+1+\delta .$$ Notice that $hI'\subset I$, hence $h(I')^{sat}\subseteq I^{sat}$ and $\operatorname{indeg}(I^{sat})\leq \delta +\operatorname{indeg}((I')^{sat})$. By Lemma \[regdim0\] we obtain $$\operatorname{reg}(I')\leq 3(d-\delta)-2-\operatorname{indeg}((I')^{sat})\leq 3d -2\delta -\operatorname{indeg}(I^{sat}) -2$$ hence $$a^0 ( {\mathcal S}_I\otimes \kappa(\pp))\leq 2d-\delta -3 -\operatorname{indeg}(I^{sat})=\nu_0 -\delta -1.$$ Finally, $\operatorname{reg}(Z_1)-d-1=\operatorname{reg}(I)-d\leq  \nu_0 -1$ by Lemma \[regdim0\].

To conclude the above case discussion, let us state a simple corollary :

\[cor:key\] Let $\pp \in \operatorname{Proj}(R)$.\
(i) If $\pi_2^{-1}(\pp)=\emptyset$ then $a^* ( {\mathcal S}_I\otimes \kappa(\pp))=\operatorname{reg}( {\mathcal S}_I\otimes \kappa(\pp))\leq \nu_0 -1$.\
(ii) If $\dim (\pi_2^{-1}(\pp))=0$ then 