This may be pretty trivial, but I can't figure it out. Suppose that $S$ is any scheme, and $f\_1:X\_1\to Y\_1$ and $f\_2:X\_2\to Y\_2$ are two morphisms of $S$-schemes, such that the closed image of each one exists. That is, there is a smallest closed subscheme $Z\_i$ of $Y\_i$ over which $f\_i$ factorizes (for $i=1,2$).


Is it true then that the closed image of $f\_1\times\_S f\_2:X\_1\times\_S X\_2\to Y\_1\times\_S Y\_2$ exists? In this case, is it equal to $Z\_1\times\_S Z\_2$ (with its natural closed immersion into $Y\_1\times\_S Y\_2$)?



Closed image of a product of morphisms
This is already false in the affine case.


The closed image of $Spec(B) \to Spec(A)$ is the spectrum of $A/K$, where $K$ is the kernel of $A \to B$. Let $A',B',K$ be analogously defined. The closed image of $Spec(B \otimes\_R B') \to Spec(A \otimes\_R A')$ is the spectrum of $(A \otimes\_R A')/L$, where $L$ is the kernel of $A \otimes\_R A' \to B \otimes\_R B'$. We have to compare this ring with $A/K \otimes\_R A'/K = (A \otimes\_R A')/\langle K,K' \rangle$. If $K=K'=0$, this asks if the tensor product of two injective rings maps is injective, which is false in general.


This discussion shows that everything is fine if, in your notation, $X\_1,Y\_2$ or $X\_2,Y\_1$ are flat over $S$.


