To what extent does the branch locus determine the covering (Chisini's conjecture)? Suppose that $X$ is a smooth projective surface over $\mathbb C$ and $f\colon X\to\mathbb P^2$ is a finite morphism branched over a curve $S\subset\mathbb P^2$. Assume in addition that all the singularities of the curve $S$ are nodes and cusps and that the covering $f$ is ``simple'' (i.e., for a general line $L\subset \mathbb P^2$, the branched covering $f^{-1}L\to L$ has only one ramification point over each $p\in S\cap L$, and the ramification index at this point is $2$). In this setting, when can one say that, for a given curve $S$, there exists at most one morphism $f$?  
I am aware of the following two results.
(a) (Kulikov, Nemirovski, 2001) The assertion is true whenever $\deg f\ge 12$;
(b) (Kulikov, 2008) The assertion is true whenever $f^*\mathcal O_{\mathbb P^2}(1)$ is very ample and $f$ is not a projection of the Veronese surface $v_2(\mathbb P^2)\subset\mathbb P^5$.
Could somebody tell me what else is known in this direction? 
Thank you in advance,
Serge
 A: I believe this is also true if $\text{deg}\ f$ equals $2$; presumably this is well-known.  To prove that $f$ is unique up to isomorphism (over $\mathbb{P}^2$), it is equivalent to prove that the $\mathcal{O}_{\mathbb{P}^2}$-algebra $f_*\mathcal{O}_X$ is unique up to isomorphism.  Of course, by Zariski's Main Theorem (or easier arguments), it suffices to prove uniqueness over the open complement $U$ of finitely many points of $\mathbb{P}^2$.  Then there is the standard short exact sequence, $$ 0 \to \mathcal{O}_{U} \xrightarrow{f^\#} f_*\mathcal{O}_X \to \mathcal{L} \to 0,$$ where $\mathcal{L}$ is an invertible sheaf on $U$ (after deleting finitely many points).  The algebra structure on $f_*\mathcal{O}_{X}$ determines an injective homomorphism of coherent sheaves, $$u:\mathcal{L}^{\otimes 2} \to \mathcal{O}_{U},$$  whose image equals the ideal sheaf $\mathcal{O}_{U}(-\underline{S})$ (again after deleting finitely many points).  Thus uniqueness of $f$ is equivalent to uniqueness of the pair $(\mathcal{L},u)$ of an invertible sheaf $\mathcal{L}$ and an isomorphism $u:\mathcal{L}^{\otimes 2}\to \mathcal{O}_{U}(-\underline{S})$ up to "equivalence", i.e., $(\mathcal{L},u)$ is equivalent to $(\mathcal{M},v)$ if there exists an isomorphism of invertible sheaves $w:\mathcal{M}\to \mathcal{L}$ such that $v\circ w^{\otimes 2}$ equals $u$.  
Of course for two such pairs, $(\mathcal{L},u)$ and $(\mathcal{M},v)$, then $\mathcal{T} := \mathcal{M}\otimes \mathcal{L}^\vee$ is an invertible sheaf and $v\otimes u^\dagger$ is an isomorphism $\mathcal{T}^{\otimes 2} \to \mathcal{O}_{U}$.  In other words, $\mathcal{T}$ is a $2$-torsion element in $\text{Pic}(U)$.  However, for every open complement $U$ of finitely many points of $\mathbb{P}^2$, $\text{Pic}(U)$ is just $\mathbb{Z}$.  Thus $\mathcal{T}$ is isomorphic to $\mathcal{O}_U$, so that the pair $(\mathcal{L},u)$ is unique up to equivalence.
I am sure the case that $\text{deg}\ f$ equals $2$ is well-known.  Just to make one observation: with the one exception where $X$ is a quadric surface, for $f$ as above, $f^*\mathcal{O}_{\mathbb{P}^2}(1)$ is not very ample.  Indeed, the pullback map $$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1))  \to H^0(X,f^*\mathcal{O}_{\mathbb{P}^2}(1))$$
is an isomorphism.  Of course, also $\text{deg}\ f \not\geq 12$.
