Assume we have two p.p. simple abelian surfaces $(A\_i,D\_i)$, i=1,2, over $\mathbb{C}$ with the following commutative diagram:


$\require{AMScd}
\begin{CD}
A\_1 @>{birational}>> A\_2\\
@V{2:1}VV @VV{2:1}V \\
A\_1/(-1) @>{birational}>> A\_2/(-1)\\
@V{\iota\_1}VV @VV{\iota\_2}V\\
\mathbb{P}^3 @>{birational}>> \mathbb{P}^3
\end{CD}$


Here the horizontal arrows, are birational maps (not morphisms) and the composition of $(2:1)$ and $\iota\_i$ is actually the morphism induced by the linear system $|2D\_i|$ for $i=1,2$. That is we have $|2D\_i|: A\_i \rightarrow \mathbb{P}^3$.


The birational map on top must be an isomorphism, since $A\_1$ and $A\_2$ are abelian. So we have $A\_1\cong A\_2$ as unpolarized abelian surfaces.


$\textbf{Question:}$ Do we actually get $(A\_,D\_1)\cong (A\_2,D\_2)$ respectively under which conditions can we conclude that we have an isomorphism of principally polarized abelian surfaces using this diagram? Or are there pairs $(A\_i,D\_i)$ where such an diagram cannot exist?


We may assume $NS(A\_i)\geq 2$ because for $NS(A\_i)=1$ we must have $(A\_1,D\_1)\cong(A\_2,D\_2)$. 


I thought one could start with the polarizations $\mathcal{L}\_i$ on $A\_i/(-1)$ coming from $\mathcal{O}\_{\mathbb{P}^3}(1)$. These pullback to $\mathcal{O}\_{A\_i}(2D\_i)$. It would be nice if one could now show that $A\_1 \xrightarrow{\cong}A\_2$ pulls $D\_2$ back to $D\_1$. This isomorphism preserves amplenes and self-intersection. So the pullback of $D\_2$ should be a principal polarization on $A\_1$ but is it $D\_1$? But for this we need to know what happens with $\mathcal{L}\_2$ under $A\_1/(-1) ---> A\_2(-1)$? But I have no idea how to get any information about this.



Behaviour of (principal) polarizations of (singular) surfaces under birational maps
@ulrich points out that the OP might be asking a question different than the one I answered, and I completely agree. So let me explain what the comment above actually proves.


Let $(A\_1,\mathcal{L}\_1)$ and $(A\_2,\mathcal{L}\_2)$ be Abelian varieties of dimension $g$ together with a specified ample invertible sheaf. Let $U\_1\subset A\_1$ and $U\_2\subset A\_2$ be dense open subsets. Let $f\_U:U\_1\to U\_2$ be an isomorphism of $k$-schemes. Let $\phi\_U:\mathcal{L}\_1|\_{U\_1}\to f\_U^\*\mathcal{L}\_2$ be an isomorphism of invertible sheaves on $U\_1$.


**Proposition.** There is a unique $k$-isomorphism $f:A\_1\to A\_2$ whose restriction to $U$ equals $f\_U$, and this extends to a $k$-isomorphism of the associated Kummer varieties. If the complement of $U\_1$ in $A\_1$ has codimension $\geq 2$ everywhere, then there exists a unique isomorphism $\phi:\mathcal{L}\_1\to f^\*\mathcal{L}\_2$ that restricts on $U$ to $\phi\_U$. There are counterexamples if the complement has codimension $1.$


**Proof.** The first result follows from what is often called "Weil extension": every $k$-rational transformation from a smooth, projective $k$-variety to an Abelian variety extends uniquely to a regular $k$-morphism on the entire smooth $k$-scheme. If the complement of $U\_1$ in $A\_1$ has codimension $\geq 2,$ then by $S2$ extension, there exists a unique extension $\phi$ of $\phi\_U$. By Krull's Hauptidealsatz, if $\text{Coker}(\phi)$ is nonzero, then the support has pure codimension $1.$ Since $\phi\_U$ is an isomorphism on $U\_1,$ then the cokernel is zero. Thus, $\phi$ is surjective. A surjective morphism of invertible sheaves is an isomorphism.


Every $g$-dimensional principally polarized Abelian variety $A$ has many ample invertible sheaves $\mathcal{L}$ with $c\_1(\mathcal{L})^{g}$ equal to $g!,$ for example the pullbacks by translations of $A$. Given two effective Cartier divisors $D\_1$ and $D\_2$ whose invertible sheaves $\mathcal{L}\_i=\mathcal{O}\_A(D\_i)$ are not isomorphic yet have $c\_1(\mathcal{L})^g$ equal to $g!,$ on the open $U\_1=U\_2=U$ equal to $A\setminus (D\_1\cup D\_2),$ there is an isomorphism $\phi\_U$ of $\mathcal{L}\_1$ and $\mathcal{L}\_2$ on $U.$ Yet $\phi\_U$ does not extend to an isomorphism $\phi.$ **QED** 


