Definition of pullback of a Weil divisor on an abelian variety? We are on an abelian variety, so Cartier divisors, Line bundles and Weil divisors are all equivalent. I would like to see the pullback of a Weil divisor. Is it true that, if $D=\sum n_i E_i$, then the pullback $\phi^* D$ is defined as $\sum n_i \phi^{-1}(E_i)$? 
If not, how is it actually defined? And are there eventually special cases in which the above definition holds?
Thanks.
 A: Pullback it's easily defined for Cartier divisors. 
Let be $\varphi : X\longrightarrow Y$ a dominant morphism of suitable varieties. Then we can show that there exists a group morphism $\varphi^* : \operatorname{Div} Y \longrightarrow \operatorname {Div} X$. The dominant hypothesis is not necessary but I'll assume it for the sake of simplicity; recall that for such a morphism, there is a field immersion $\varphi^* :k(Y)\hookrightarrow k(X)$, where $k(-)$ denotes the field of rational functions. 
First of all, you should know that a Cartier divisor $D$ over $Y$ can be represented by a couple $(\{U_i\}_{i\in I},\{f_i\})$ where $\{U_i\}$ is an open cover of $Y$ and $f_i$ is a rational non zero function over $Y$, such that $f_if_j^-1$ is a regular function over $U_i\cap U_j$, for all $i,j\in I$. 
With such identification, we define 
$$\varphi^* D\in\operatorname{Div}(X)$$
by the means of the couple
$$(\{\varphi^{-1}(U_i)\}_{i\in I}, \{\varphi^* f_i\})$$
where $\varphi^*$ is the forementioned field extension.
It's straightforward to check that $\varphi^*D$ is a Cartier divisor.
