Inverse image of dominating component Let $f:X\rightarrow Y$ be a surjective proper morphism of varieties, with $Y$ irreducible. The fibers of $f$ are irreducible (not necessarily of same dimension). Let $U$ be an irreducible, dense, open subset of $Y$, such that $f^{-1}(U)$ is isomorphic to $U$.
Then $f^{-1}(U)$ is an irreducible open set of $X$. Is $\overline{f^{-1}(U)}$ an irreducible component of $X$. It is clearly an irreducible closed set of $X$. Why is it maximal?
If $g:Y\rightarrow Z$ is again a proper surjective morphism with $Z$ irreducible. When can we say that $\overline{f^{-1}(U)}$ is the only component of $X$ dominating $Z$?

Note for the following that making a statement about $h=g\circ f$ doesn't really differ from making a statement about $f$, because $h$ is also surjective and proper with an irreducible target, and that's all we know about $f$. The following also answers the question about $f$.
As per request in the comment, we will prove the following statement: Let $f\colon X\to Y$ be a proper surjective morphism of varieties with $Y$ irreducible. Then, there is only one irreducible compoent of $X$ dominating $Y$ if and only if there is a nonempty open subset $U\subseteq Y$ such that $f^{-1}(U)$ is irreducible.
Assume first there is an $U$ with this property and let $W\subseteq X$ be any irreducible component of $X$ dominating $Y$. Since $f(W)=Y$, we know that $W\cap f^{-1}(U)$ is nonempty and open in $W$. It follows that 
$$\overline{f^{-1}(U)}\supseteq\overline{W\cap f^{-1}(U)}=W,$$ 
the equality holding because $W$ is irreducible.
By maximality of $W$ on the other hand, $W=\overline{f^{-1}(U)}$, so there is only one component of $X$ dominating $Y$ and it is equal to $\overline{f^{-1}(U)}$.
For the converse, assume that there is only one component of $X$ which dominates $Y$, call it $W$. Then, $f(W)=Y$. Let $H\subseteq X$ be the union of all other irreducible components of $X$, it is a closed subset of $X$ and $f(H)$ is a proper, closed subset of $Y$. Set $U:= Y\setminus f(H)$, this is a nonempty open subset of $Y$. Furthermore, $f^{-1}(U)$ is a nonempty open subset of $W$ by construction, hence $W=\overline{f^{-1}(U)}$. Since $W$ is irreducible, so is $f^{-1}(U)$.
