I'm studying some stuff about Picard groups and divisors and I started wondering something, here it is:
Let $X$, $Y$ be algebraic varieties and $f:X\rightarrow Y$ a bimeromorphic map between them. Let $D\subset X$ and $D'\subset Y$ be divisors and $[D]\in H^{1,1}(X,\mathbb{Z})$, $[D']\in H^{1,1}(Y,\mathbb{Z})$ their fundamental classes.
Suppose i want to prove that $f$ induces a bimeromorphic map between $D$ and $D'$ and that the open sets in which $f$ is biholomorphic have complements of codimention at least 2, so i don't have problems when restricting to $D$ and $D'$.
My question is: is it enough to prove that $f^*([D'])=[D]$? I.e. at a homological level, if $f^*$ brings the class $D'$ to the class of $D$ is it enought to say that $f$ maps $D$ bimeromorphically into $D'$?
Is the situation is different if $X$ and $Y$ are K3 surfaces, so then $Pic(X)=H^{1,1}(X,\mathbb{Z})$ and $Pic(Y)=H^{1,1}(Y,\mathbb{Z})$ and i can think in terms of classes inside Picard groups?
Finally what happens if $f^*([D'])=-[D]$? what does that minus sign represent geometrically? that i cannot have the bimeromorphic map?
This is false when $X = Y = \mathbb P^2$, and $f$ is the identity. Take $D$ to be one line and $D′$ a different line. These have the same cohomology class but $f$ doesn't induce a map between them.
On the other hand, I think there are some cases when it is true. Suppose that $D$ and $D'$ are both the unique divisors in their cohomology classes (for example, they are irreducible curves of negative self-intersection on a surface). The divisor $f^{-1}(D')$ is certainly in the cohomology class $f^*([D'])$. If the only divisor in that class is $D$, then it must be that $f^{-1}(D') = D$ and you have the map you want. This can occur in interesting cases: for example, on a K3 surface with infinitely many $(-2)$ curves, to conclude that some automorphism sends one to another it's enough to check this at the numerical level.
It's impossible to have $f^*([D]) = -[D']$, at least if $X$ is projective. $f^*([D])$ and $[D']$ are both effective divisor classes, but on a projective variety the (pseudo)effective cone has to be strongly convex (meaning it never contains both a class and its opposite). This follows pretty easily by cutting by general hyperplane sections to reduce to the 2-dimensional setting and then invoking the Hodge index theorem -- offhand I don't know a reference.