Let $\varphi \colon X\to S_2$ be a proper dominant morphism between varieties. (This varieties has the same dimension in my example if it helps) I have another morphism $\psi\colon S_1\to S_2$ which is flat. $S_1$ is not necessary irreducible, it may has several components, but all of these component have the same dimension. Let $\varphi_\psi \colon X\times_{S_2}S_1\to S_1$ be the base change of $\varphi$.
Is it true that for any irreducible component of $X\times_{S_2}S_1$, its image under the map $\varphi_\psi$ is dense in some irreducible component of $S_1$?
It is not difficult to show that $\varphi_\psi$ is dominant. But why $S\times_{S_2}S_1\to S_1$ cannot have some components which are "contracted" under the map $\varphi_\psi$?
Here is my own answer.
Let $Y=X\times_{S_2}S_1$ and $D$ be some irreducible component of $Y$. The image $\psi_\varphi(D)$ is dense in $X$ as $\psi_\varphi$ is flat and flat morphism is open. As $\varphi$ is dominant $\varphi(\psi_\varphi(D))$ is dense in $S_2$. This means that if we restrict everything to some open subset of $U\subset S_2$ then $D$ is restricted to non-empty open subset $Z_U$ of $Z$.
Now general flatness says that over some open $U\subset S_2$ the morphism $\varphi$ is flat, so we can assume $\varphi$ to be flat. But in this case $\varphi_\psi$ is flat and hence open. The statement follows.