Let $\mathcal{C}$ be a category in which every fiber product exists. Let $S' \rightarrow S$ be a morphism in $\mathcal{C}$. Let $(\mathcal{C}\downarrow S)$ and $(\mathcal{C}\downarrow S')$ be the slice categories. Let $F\colon (\mathcal{C}\downarrow S) \rightarrow (\mathcal{C}\downarrow S')$ be the functor defined by $F(X) = X\times_S S'$. Does $F$ preserve limits?
Does fiber product preserve limits?
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category-theory
2 Answers
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Yes. In fact, $(-) \times_S S'$ preserves limits because it is the right adjoint of the functor that takes an object $X \to S'$ in $(\mathcal{C} \downarrow S')$ to the object $X \to S' \to S$ in $(\mathcal{C} \downarrow S)$.
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0I understand how you and Makoto Kato are definng F, but I have a question: is your notation really correct? The way you write it $(-) \times_S S'$ appears to be a functor in $\mathcal{C}$ and not on the slice, and can be misleading, as shown below in Augusti Roig question/comment – 2012-12-27
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0How could it possibly be a functor on $\mathcal{C}$? There is no chance of confusion. – 2012-12-27
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0The source of my puzzlement is this: $F$ is a functor on a slice cat, whose objects are really morphisms $f$ in $\mathcal{C}$. So - by writing the functor $F$ in that way - you end up with expressions like $f \times_S S'$ where you have a fiber product between a morphism $f$ and an object $S'$. I wonder : is this standard notation? if yes, where? – 2012-12-27
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0The notation we use is completely standard in algebraic geometry, for example, and fairly common elsewhere. It is an accepted abuse of notation. – 2012-12-27
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0Thank you. Could you please suggest an algebraic geometry text where these functors/arguments are used? – 2012-12-27
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0I don't really know what you're asking. This argument can be found in various category theory textbooks – certainly in Awodey (Ch. 9). – 2012-12-27
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0Indeed it is in Awodey's text, but I was under the impression that this functor F is used in algebraic geometry. So I am asking for an algebraic geometry text, if it is possible. Or, more in general, a good introductory text on algebraic geometry which uses categorical language – 2012-12-27
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0This functor is used whereever Grothendieck's relative point of view is fruitful: algebraic geometry is just one of them. There is literally no algebraic geometry content to this question or its answer. – 2012-12-27
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Yes and another explanation: your functor, being itself a limit, commutes with limits.
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0True, but _this_ functor is not a limit. The functor $(\mathcal{C} \downarrow S) \times (\mathcal{C} \downarrow S) \to (\mathcal{C} \downarrow S)$ that sends $(X \to S, Y \to S)$ to $X \times_S Y$ _is_, and it is right adjoint to the diagonal functor $\Delta : (\mathcal{C} \downarrow S) \to (\mathcal{C} \downarrow S) \times (\mathcal{C} \downarrow S)$. – 2012-12-27