Do disjoint unions and fiber products commute?
In other words, is the following statement true?
Statement: Let $C$ be a category with (infinite) coproducts and fiber products. Let {$U_{i}$} be a family of objects in $C$, and denote the coproduct of them by $U = \coprod_{i}U_{i}$. Moreover, let $U_{i} \to X$ and $Y\to X$ be morphisms in $C$, and $U\to X$ be the morphism induced by the universality of coproduct. Then, $U\times_{X}Y \cong \coprod_{i}(U_{i}\times_{X}Y)$.
If $C$ is the category of schemes, this statement will be true. This is because fiber products of schemes are constructed locally at first, and glued together.
However, I could not prove this by using universality (i.e. in categorical settings).
My questions are:
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Is the above statement true? If so, then how can one prove it?
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If the statement is false, what kind of counter example exists?
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If the statement is false, then, please change the statement replacing "coproducts" by "disjoint unions". Is the NEW statement true?
Here, disjoint union of $U_{i}$'s means coproduct $U=\coprod_{i}U_{i}$ satisfying that the fiber products $U_{i}\times_{U}U_{j}$ are the strict initial object if $i \neq j$. Here, strict initial object means initial object $\phi$ such that for any object $X$, the set of morphisms $Hom(X, \phi)$ is the empty set if $X$ is not isomorphic to $\phi$. (This is the generalization of empty set in the category of sets or schemes.)
Later
Counterexamples for the first statement exist (e.g. the category of pointed sets or the opposite of the category of sets).
However, those are not for the refined statement in my question 3.
Does anybody have ideas for it?
Best Answer
Answer to 2: If $A$, $B$, $C$ are three sets, it is not true in general that $(A\times B)\coprod C=(A\coprod C)\times (B\coprod C)$. Hence the category $(Sets)^{op}$ is a counterexample.
EDIT: (March 2021) If we allow infinite coproducts, the category of affine schemes is a counterexample with disjoint unions (in the sense of question 3): take $P=$ the set of prime numbers, $U_p=\mathrm{Spec}(\mathbb{F}_p)$ for $p\in P$, $X=\mathrm{Spec}(\mathbb{Z})$, and $Y=\mathrm{Spec}(\mathbb{Q})$. Each $U_p\times Y$ is empty, but $(\coprod_p U_p)\times Y$ is not: $A:=\prod_p\mathbb{F}_p$ is not a torsion ring, so $A\otimes\mathbb{Q}\neq0$.