When does a generic point map to a generic point? This question may be too vague, so feel free to specialize to particular examples.
Given a morphism of schemes $f:X\to Y$, I want to know what conditions one can impose on $f,X$ or $Y$ such that a generic point of $X$ will map to a generic point of $Y$. For example, if $X$ and $Y$ are irreducible etc.
 A: Let $f:X\rightarrow Y$ be a map between irreducible schemes with generic points $\eta$ and $\theta$ respectively. It's possible to prove that $f(\eta)=\theta$ is equivalent to $f$ being dominant without assuming that the schemes are reduced. Here is the proof:
If $f(\eta)=\theta$ we have $\overline{f(X)}\supseteq \overline{f(\eta)}=\bar{\theta}=Y$. On the other hand, if $f$ is dominant then, as $f(\overline{A})\subseteq \overline {f(A)}$ for every continuous function and every set $A$, we have $$f(X)=f(\bar{\eta})\subseteq \overline{f(\eta)}.$$
So $f(\eta)$ is a point whose closure is $Y$, hence $f(\eta)=\theta$.
A: Only irreducible sets have a unique generic point. Since you mention 'the' generic point I will assume $X,Y$ are irreducible from now on.  Let $\eta$ be the generic point of $Y$ and $\mu$ of $X$. Then $\eta$ is contained in each affine open of $Y$ and the same goes for $\mu$ in $X$. 
So we can restrict our attention to some affines $U\subset X$ and $V \subset Y$.
The map becomes
$$f: U = \mbox{Spec } R \rightarrow \mbox{Spec } S = V,$$
$$\varphi: S \rightarrow R$$
and your question reduces to $\varphi^{-1}(\sqrt{(0)}) = \sqrt{(0)}$. Being irreducible gives that this nilradicals are prime. If you further assume that the schemes are reduced, then the nilradicals will be zero. Then we demand injectivity of $\varphi$. This must hold for every open affine $V$ and $U$, which is equivalent to the sheaf component $f^{\#}$ being injective or $f$ being dominant, i.e. having dense image.
