# Does fiber product preserve limits?

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?

-

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

-
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.

-
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