Components of a Fiber Product Suppose $X$, $Y$, and $Z$ are smooth irreducible schemes [EDIT: of finite type over an algebraically closed field of characteristic zero], and $X \to Z$ and $Y \to Z$ are dominant maps.
I have a certain point $(x, y)$ in the fiber product $X \times_Z Y$; and I'd like to know that it lies in some component of the fiber product which dominates $Z$.
What condition on the differentials $df|_x : T_x X \to T_z Z$ and $df|_y : T_y Y \to T_z Z$ will allow me to verify this?
It's fairly easy to see that the surjectivity of $df|_x$ (or $dg|_y$) alone is enough to guarantee the desired conclusion --- I'm hoping here for something weaker that would still suffice. I'd by happy if the surjectivity of $df|_x + dg|_y$ suffices, for example; but if it does not, I would be interested in any other weaker condition too.
 A: [Observation of David Yang.]
The condition $df|_x(T_x) + dg|_y(T_y) = T_z$ is not sufficient; more generally, the only closed condition depending on the images of $df|_x(T_x)$ and $dg|_y(T_y)$ that could be sufficient is the surjectivity of either $df|_x$ or $dg|_y$.
To see this, it suffices to construct a counterexample where $df|_x(T_x)$ and $dg|_y(T_y)$ are codimension one subspaces in general position. For this, we can take $Z = \mathbb{P}^n$, and $X = Y = \text{Bl}_\Lambda \mathbb{P}^n$, where $\Lambda$ is a subspace of codimension two.
A: Edit: I gave a proof of a false statement earlier. It can be salvaged to the folowing much weaker statement: the point $(x,y)$ is in the dominant irreducible component if we have $df|_x(T_x)+dg|_y(T_y)=T_z,$ together with the following assumption: 
Let $r=\operatorname{rank}(df|_x).$ Then there is a local coordinate system $z_1,\ldots, z_n$ with origin $z\in Z$ such that
(1) the first $r$ coordinates $z_1,\dots, z_r$ define a full-rank map $(f_1,\dots, f_r):X\to A^r$ and 
(2) for fixed $(z_1,\dots, z_r)$, let $X_{(z_1,\dots, z_r)}=(f_1,\dots, f_r)^{-1}(\{(z_1,\dots, z_r)\})$ be the fiber. I ask that the map $(f_{r+1},\dots, f_n):X_{(z_1,\dots, z_r)}\to A^{n-r}$ be dominant for any tuple $(z_1,\dots, z_r)$ in a neighborhood of 0.
Don't know how useful this is. But in this case one can see using the implicit function theorem that the intersection $f(D_x)\cap g(D_y)$ of small disks near $x,y$ has nonempty interior, and the statement follows.
