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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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