Let $X\to Z$ and $Y\to Z$ be scheme morphisms (everything noetherian, of finite type over an algebraically closed field $k$ of char $0$), and consider a morphism to another scheme $X\times_ZY\to T$ that is injective. Assuming that for all $x\in X$, and $y\in Y$, the restrictions $\{\mathrm{Spec}(k(x))\}\times_ZY\to T$ and $X\times_Z\mathrm{Spec}(k(y))\to T$ are embeddings, can we conclude that $X\times_Z Y\to T$ is an embedding?
I have a hunch this is false, but maybe under good circumstances?
EDIT: I can assume that $X\times_Z Y$ is the proj of (the symmetric product of) a coherent sheaf over $X$.
That is not true. Consider the map $$\mathbb C^2 \ni (s,t) \mapsto (s, t^2, t^3) \in \mathbb C^3.$$ That map is injective, and has injective differential everywhere except in zero. There the kernel is generated by the tangent vector $\frac \partial {\partial t}$. However, if we make the change of coordinates $$u = s+t, v = s-t,$$ then both tangent vectors $\frac \partial{\partial u}$ and $\frac \partial {\partial v}$ are not in the kernel. So the slices $\{u=0\}$ and $\{v=0\}$ are embedded as closed immersions into $\mathbb C^3$.