Injective maps on cohomology and Kahler manifolds Compact Kahler manifolds have the property that surjective maps induce injections on cohomology with coefficents in $\mathbb{Q}$ (That is, if $X,Y$ compact Kahler, then a surjective map $\phi: X \rightarrow Y$ induces injections $\phi^*: H^i(Y, \mathbb{Q}) \rightarrow H^i(X, \mathbb{Q})$ for all $i$, [Voisen, Hodge Theory I, p 177]).
Question: I'm wondering if I should think of this as a property of compact Kahler manifolds, or as an instance of something more general. For example, can the Kahler condition be replaced with a more general class of manifolds (not dropping the compactness hypothesis). Perhaps one that includes not just complex manifolds but maybe a few manifolds of odd (topological) dimension? I know that if we require that $\dim X = \dim Y$ then the fact above is true more generally just for compact oriented manifolds, for formal reasons.
 A: Suppose one has compact symplectic manifolds $(X,\omega), (Y,\sigma)$, and a map $f:X\to Y$ such that $f^\ast\sigma=\omega$ (if $f$ is also a diffeomorphism, then this map is a symplectomorphism, but I'm not sure the terminology in this case). Then $f^\ast: H^*(Y;\mathbb{Q})\to H^\ast(X;\mathbb{Q})$ will be injective. If $dim Y=2k$, then $\sigma^k\neq 0\in H^{2k}(Y)$ is a fundamental class, and $dim X=2n \geq 2k$, then $\omega^n\neq 0\in H^{2n}(X)$. So $f^\ast(\sigma^k)= \omega^k \neq 0 \in H^{2k}(X)$. 
Now, consider $\alpha \in H^j(Y)$. By Poincare duality, there exists $\beta\in H^{2k-j}(Y)$ such that $\alpha\cup \beta = \sigma^k$. Then $f^\ast(\alpha\cup \beta)=\omega^k\neq 0$, so $f^\ast \alpha\neq 0$. 
A: In order to have this property it is sufficient to require that $\phi^{-1}(y)$ is a non-zero cycle in $H_*(X,\mathbb Q)$, where $y$ is a generic point in $Y$. This holds indeed when $X$ and $Y$ are Kahler.
Let me give an example showing that this does not work when $X$ and $Y$ are just complex. 
Example. Let $X$ be a Hopf surface $X=(\mathbb C^2\setminus 0)/\mathbb Z$ where $\mathbb Z$ is acting on $\mathbb C^2$ by multiplication by (say) $2$. Then there is a fibration $\phi\colon X\to \mathbb CP^1=Y$. This fibration comes  from the standard $\mathbb C^*$-action on $\mathbb C^2$. Now  $\phi^{-1}(y)$ is null-homlogous.
