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:

Is the above statement true? If so, then how can one prove it?

If the statement is false, what kind of counter example exists?

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?

strictnessof the initial object considered part of the standard definition? I am inclined to think not: that strictness of the initial object is considered more of a distributivity condition (it is equivalent to asking that $X \times -$ preserve the initial object). For example, in my way of thinking, $Vect$ has disjoint coproducts. $\endgroup$ – Todd Trimble♦ Feb 25 '12 at 2:49andthat each coproduct injection is monic (i.e. that $U_i \times_U U_j$ is $U_i$ when $i=j$). If one adds to this the assumption that coproducts are universal (your question 1), one obtains the notion of anextensive categorynlab.mathforge.org/nlab/show/extensive+category . It is interesting to note that the assumption that binary coproducts are universalimpliesthat the initial object is strict. $\endgroup$ – Mike Shulman Feb 25 '12 at 6:15alwaysthe case; it requires nothing about coproducts or extensivity. More generally, if $(U_i \to X)$ is UEE and $(V_j \to X)$ is a family such that every $U_i$ factors through some $V_j$, then $(V_j \to X)$ is UEE. See, for instance, C2.1.6 inSketches of an Elephantvol 2. $\endgroup$ – Mike Shulman Mar 1 '12 at 0:33