Let $X$ be a projective (not necessarily smooth) normal variety of general type over $\mathbb{C}$. Let $A$ be an abelian variety and let $A\to X$ be a surjective morphism. 



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I have the feeling that one can pull-back differential forms on a resolution $\tilde{X}$ of ${X}$ (by which I mean sections of $\omega\_{\tilde{X}}^{\otimes n}$) to obtain differential forms on $A$. But how does one do that?



Can an abelian variety dominate a variety of general type?
Theorem: Let $A$ be an abelian variety and $f:A\to X$ a dominant morphism to a projective variety of general type, then $\dim X=0$. 


Proof: Replacing $f$ by a birational model of the Stein factorization we may assume that $f':A'\to X'$ is a projective morphism of smooth varieties with $f'\_\*\mathcal O \_{A'}=\mathcal O \_{X'}$, $X'$ is of general type and $A'\to A$ is birational (and hence $\kappa (A')=\kappa (A)=0$). Let $F$ be a general fiber, then $\kappa (F)\geq 0$ (since $K\_{A'}|\_F\cong K\_F$). By Theorem 3 of <http://www.numdam.org/article/CM_1981__43_2_253_0.pdf> we have $\kappa (A')\geq \kappa (F)+\kappa(X')$. But then $\dim X'=\kappa(X')=0$.


