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What does it exactly mean to say that in a certain category pushouts and pullbacks "commute"? Is it the same to say that they "distribute"?

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3 Answers 3

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The condition Martin and Todd mention is indeed a sort of distributivity condition. It is also often called {\em stability} of pushouts. I think that it should not be called commutativity.

Let D and C be small categories, and A a category with D-shaped limits and C-shaped colimits. Then D-limits commute in A with C-shaped colimits when the functor $[D,A]\to A$ which calculates the limit preserves C-colimits. This is equivalent to saying that the functor $[C,A]\to A$ which calculates the colimit preserves D-limits. The most famous example is commutativity of finite limits with filtered colimits.

Note, however, that commutativity of pullbacks and pushouts in this sense is rare, and is probably not what was meant.

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Welcome, Steve! –  David Roberts Nov 17 '10 at 23:15
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If we pass from category theory to homotopy theory, we can consider whether D-shaped homotopy limits commute (up to natural weak equivalence) with C-shaped homotopy colimits. This is not so rare. If the category is pointed then it is the same as what is commonly called stable. Typical examples are spectra, or (unbounded) chain complexes of R-modules. It's also the same as saying that a square diagram is a homotopy pushout iff it is a homotopy pullback. Also the same as: finite homotopy colimits commute with all homotopy limits (or vice versa). –  Tom Goodwillie Nov 18 '10 at 3:06
    
Good point about the homotopy case. One example (in the non-homotopy case) where pushouts and pullbacks do commute is in a groupoid. But pullbacks and pushouts in groupoids are not very interesting. –  Steve Lack Nov 18 '10 at 10:40
    
Thank you for the answer, Steve. May I formulate it in this way: The commuting property is fulfilled iff for every functor $F:C\times D\to A$ it does not matter if I first apply the colimit functor to the first "factor" and get a functor $F':D\to A$ to which I apply the limit functor or if I do it the other way round? –  roger123 Nov 18 '10 at 14:10
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Yes, that's it. –  Steve Lack Nov 19 '10 at 1:28

I think one would need some context to be sure how this is meant to be interpreted, but one possible interpretation would be subsumed under the condition that a pullback functor between slice categories preserves colimits.

So: suppose we have a pushout $h: B \to P \leftarrow C: j$ of a diagram

$$B \stackrel{f}{\leftarrow} A \stackrel{g}{\to} C$$

and suppose given an arrow $k: Q \to P$. Then you can pull back the pushout diagram targeted at $P$ along $k$ to get a square terminating at $Q$. If this square is a pushout, then you could say that the pullback functor $k^\ast$ preserves the pushout.

This happens for example in toposes $E$: every pullback functor

$$k^\ast: E/P \to E/Q$$

preserves colimits.

(Beaten by about a minute by Martin!)

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Basically all compatibility conditions may be phrased in form of commutativity; in particlar distributivity.

Pushouts commute with pullbacks, if the following holds: For all morphisms $B \to A,C,D \to R$ the canonical morphism

$(A\times_R D) \coprod_B (C \times_R D) \to (A \coprod_B C) \times_R D$

is an isomorphism. And indeed this looks like the distributivity law $AD+CD = (A+C)D$.

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Thank you, Martin. You assert in particular that "commute" and "distribute" mean the same in the context of the question, right? –  roger123 Nov 16 '10 at 18:21
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I don't understand. Do D and A+C have maps to R? –  Tom Goodwillie Nov 18 '10 at 6:03
    
@Tom: Yes. A,C,D have maps to R and by the universal property there is a uniquely determined map from A+C to R which fits into the usual commutative diagram. –  Johannes Hahn Nov 18 '10 at 11:23
    
(I've made an edit) –  Martin Brandenburg Nov 18 '10 at 11:51

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