Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 7 down vote accepted

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 $(\mathcal{C} \downarrow S)$.

share|improve this answer
    
I 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 –  magma Dec 27 '12 at 10:20
    
How could it possibly be a functor on $\mathcal{C}$? There is no chance of confusion. –  Zhen Lin Dec 27 '12 at 10:43
    
The 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? –  magma Dec 27 '12 at 11:53
    
The notation we use is completely standard in algebraic geometry, for example, and fairly common elsewhere. It is an accepted abuse of notation. –  Zhen Lin Dec 27 '12 at 11:58
    
Thank you. Could you please suggest an algebraic geometry text where these functors/arguments are used? –  magma Dec 27 '12 at 12:43

Yes and another explanation: your functor, being itself a limit, commutes with limits.

share|improve this answer
    
True, 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 $\Delta : (\mathcal{C} \downarrow S) \to (\mathcal{C} \downarrow S) \times (\mathcal{C} \downarrow S)$. –  Zhen Lin Dec 27 '12 at 4:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.