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:

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

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

  3. 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.)


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?

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    $\begingroup$ Is your notion of disjoint coproduct/union standard in the categorical literature? More exactly, is strictness of 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 2:49
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    $\begingroup$ The standard notion of "disjoint coproduct" in the categorical literature asks that $U_i \times_U U_j$ is a (not assumedly strict) initial object for $i\neq j$, and that 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 an extensive category nlab.mathforge.org/nlab/show/extensive+category . It is interesting to note that the assumption that binary coproducts are universal implies that the initial object is strict. $\endgroup$ – Mike Shulman Feb 25 '12 at 6:15
  • $\begingroup$ Thanks for your comments. My definition of disjoint union is the one in "Algebraic Spaces" written by Donald Knutson. And I was interested in the problem that if U_i -> X is a universal effective epimorphic family, then, the induced map U -> X is a universal effective epimorphism or not. I thought in order to prove this, one needs properties like in my questions, but now I am a little confused. Is this statement true or not? $\endgroup$ – Hiro Feb 25 '12 at 8:54
  • $\begingroup$ "the induced map U -> X is a universal effective epimorphism or not." It should be true - I have an inkling that you are dealing with a superextensive site. To focus Mike's link a little, see nlab.mathforge.org/nlab/show/… $\endgroup$ – David Roberts Feb 27 '12 at 4:13
  • $\begingroup$ The question you say you were interested in is actually always the 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 in Sketches of an Elephant vol 2. $\endgroup$ – Mike Shulman Mar 1 '12 at 0:33

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.

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  • $\begingroup$ Thanks for your comment. Your counter example is interesting... How about question 3? Do you have any idea? $\endgroup$ – Hiro Feb 24 '12 at 22:39

The more general question is whether pullbacks and colimits commute. This is true if the category $\mathcal C$ is locally cartesian closed, i.e. every slice category $\mathcal C /c$ of objects over a given object $c$ of $\mathcal C$ is cartesian closed. For then pullback has a right adjoint.

These ideas are of interest in the topological case, though even if $\mathcal C$ is a convenient category of spaces, the categories $\mathcal C/c$ are not always cartesian closed.

Nonetheless, the use of such categories was initiated by R. Thom and continued by Peter Booth, see for example:

Booth, Peter I. A unified treatment of some basic problems in homotopy theory. Bull. Amer. Math. Soc. 79 (1973), 331–336.

and his related papers: these ideas were also taken up by Ioan James under the term "Fibrewise topology", see

Crabb, Michael; James, Ioan; Fibrewise homotopy theory. Springer Monographs in Mathematics. Springer-Verlag London, Ltd., London, 1998.

There is a paper

Johnstone, P. T. On a topological topos. Proc. London Math. Soc. (3) 38 (1979), no. 2, 237–271.

giving a locally cartesian closed category, in fact a topos, with sequential spaces as a reflective subcategory, but this has not yet been used in algebraic opology, to my knowledge.

August 19, 2014 A doctoral thesis in this area, "Topos Theoretic Methods in General Topology" by Hamed Harasani, Bangor 1988. is available here.

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It's true for the category of sets and various related categories, such as topological spaces. It's false for pointed sets. And, as Laurent Moret-Bailly has pointed out, it's false for the opposite of the category of sets.

When it's true of a particular category, it may be because for each $Y\to X$ the pullback functor $U\mapsto U\times_XY$ has a right adjoint -- in which case pullbacks preserve all colimits, not just coproducts.

EDIT Take the category of commutative semigroups. (An object is a set with a commutative and associative addition law.) Categorical product is the obvious thing. The initial object (empty set) is strict in your sense. The coproduct of $A$ and $B$ is the disjoint union of $A$, $B$, and $A\times B$ with the obvious addition law. Product is not distributive over coproduct.

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  • 1
    $\begingroup$ Thank you for your comment! Your examples are exactly counterexamples for my statement. But Those are not for the NEW statement since the category of pointed sets has no "strict" initial object. Do you have any idea for the question 3? $\endgroup$ – Hiro Feb 24 '12 at 22:59
  • $\begingroup$ Is it true that in any (grothendieck?) topos pullback has a right adjoint? (and hence fibre products commute with disjoint unions) $\endgroup$ – Yosemite Sam Feb 25 '12 at 15:08
  • $\begingroup$ Yes, I believe so. $\endgroup$ – Tom Goodwillie Feb 25 '12 at 18:42
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    $\begingroup$ In fact, the condition that "coproducts are disjoint and universal" is part of a characterization of toposes due to Giraud: see SGA 4, IV, Theorem 1.2. $\endgroup$ – Laurent Moret-Bailly Feb 25 '12 at 20:57
  • $\begingroup$ To see that the pullback functor $(\mathcal C\downarrow Y)\to (\mathcal C\downarrow X)$ induced by a map $X\to Y$ in a topos has a right adjoint: First use that the category of objects over $Y$ is again a topos. This reduces you to the case when $Y$ is terminal. Now pullback $\mathcal C\to (\mathcal C\downarrow X)$ is given by $Z\mapsto (Z\times X\to X)$. Use the existence of exponentials. A right adjoint is given by $(W\to X)\mapsto lim(W^X\rightarrow X^X\leftarrow \star)$. $\endgroup$ – Tom Goodwillie Feb 26 '12 at 16:46

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