nLab
fiber product

A fibre product or fiber product is simply a product in a slice category. The fibre product of two morphisms is the same as their pullback; accordingly, a fiber product more than two morphisms is often called a wide pullback.

More explicitly, for $f:A\to C$ and $g:B\to C$ two morphisms in a category $C$, the fiber product $A{\times }_{C}B$ of $A$ with $B$ over $C$ is, if it exists, the pullback

$$\begin{array}{ccc}A{\times }_{C}B& \to & B\\ \downarrow & & {\downarrow }^g\\ A& \stackrel{f}{\to }& C\end{array}.$$\array{ A \times_C B &\to& B \\ \downarrow && \downarrow^g \\ A &\stackrel{f}{\to}& C } \,.

This term comes from thinking of $A$ and $B$ as bundles over $C$; then the fiber of $A{\times }_{C}B$ over a generalized element $x$ of $C$ is the product of the fibers of $A$ and $B$ over $x$. In other words, the fiber product is the product taken fiber-wise.

Of course, the fiber of $A$ at the generalized element $x:I\to C$ is itself a pullback $I{\times }_{C}A$; the terminology depends on your point of view.

You're English, David, so how come you don't like ‘fibre’? —Toby

I just thought it was policy here. That’s what the HowTo page says anyway. —David

That's only for page titles. (I know, because I wrote it!) Actually, now that we have a usable system of redirects, even that may not be so important. The English Wikipedia does fine with a policy only of including redirects for alternate spellings. —Toby

Examples

In $C=$ Set the fiber product is given by the usual formula

$$A{\times }_{C}B=\{(a,b)\in A\times B\mid f(a)=g(b)\}.$$A \times_C B = \left\{ (a,b) \in A \times B \;|\; f(a) = g(b) \right\} \,.

Chris: I’d appreciate it if someone could add an explanation of fiber products as limits of compsimplicial diagrams $A\times B,A\times C\times B,A\times C\times C\times A,...$, making the coface maps explicit.

Urs: I think it should work like this:

the first two coface maps are in full detail given by

$$A\times B\stackrel{({\mathrm{Id}}_A\circ p_1)\times (f\circ p_1)\times ({\mathrm{Id}}_B\circ p_2)}{\to }A\times C\times B$$A\times B \stackrel{(Id_A \circ p_1) \times (f\circ p_1) \times (Id_B \circ p_2)}{\to} A \times C \times B

and

$$A\times B\stackrel{({\mathrm{Id}}_A\circ p_1)\times (g\circ p_2)\times ({\mathrm{Id}}_B\circ p_2)}{\to }A\times C\times B$$A\times B \stackrel{(Id_A \circ p_1) \times (g\circ p_2) \times (Id_B \circ p_2)}{\to} A \times C \times B

With that it is clear that the fiber product $A{\times }_{C}B$ is the equalizer of these two maps, i.e. the limit over the diagram

$$A\times B\overrightarrow{\to }A\times C\times B.$$A \times B \stackrel{\to}{\to} A \times C \times B \,.

I am guessing, but haven’t really checked in detail that the further coface maps continue this: the “inner” ones come from the diagonal $C\to C\times C$ and the outer ones from $f$ and $g$ as above. So for instance the next three would be

$$A\times C\times B\stackrel{({\mathrm{Id}}_A\circ p_1)\times (f\circ p_1)\times ({\mathrm{Id}}_C\circ p_2)\times ({\mathrm{Id}}_B\circ p_3)}{\to }A\times C\times C\times B$$A \times C \times B \stackrel{(Id_A \circ p_1) \times (f \circ p_1) \times ( Id_C\circ p_2) \times (Id_B \circ p_3)}{\to} A \times C \times C \times B

and

$$A\times C\times B\stackrel{({\mathrm{Id}}_A\circ p_1)\times ({\mathrm{Id}}_C\times {\mathrm{Id}}_C\circ p_2)\times ({\mathrm{Id}}_B\circ p_3)}{\to }A\times C\times C\times B$$A \times C \times B \stackrel{(Id_A \circ p_1) \times (Id_C \times Id_C \circ p_2) \times (Id_B \circ p_3)}{\to} A \times C \times C \times B

and

$$A\times C\times B\stackrel{({\mathrm{Id}}_A\circ p_1)\times ({\mathrm{Id}}_C\circ p_2)\times (g\circ p_3)\times ({\mathrm{Id}}_B\circ p_3)}{\to }A\times C\times C\times B$$A \times C \times B \stackrel{(Id_A \circ p_1) \times (Id_C \circ p_2) \times ( g \circ p_3) \times (Id_B \circ p_3)}{\to} A \times C \times C \times B

This way, unless I am making a mistake, the ordinary limit over the diagram

$$A\times B\overrightarrow{\to }A\times C\times B\overrightarrow{\overrightarrow{\to }}A\times C\times C\times B$$A \times B \stackrel{\to}{\to} A \times C \times B \stackrel{\to}{\stackrel{\to}{\to}} A \times C \times C \times B

would still be $A{\times }_{C}B$, but the homotopy limit would pick up the right first degree correction for the homotopy fiber product.

Chris: Thanks, Urs. I’ll continue thinking about this. While I was at the coffee shop, I realized that to form $A{\times }_{C}B$ in a homotopically meaningful way, we should somehow resolve $B$. So if $𝒟$ is our category and ${𝒟}_C$ is the category of objects over $C\in 𝒟$, then we have an adjoint pair of functors $R:=C\times ?:𝒟\to {𝒟}_C$ and $L:={𝒟}_C\to 𝒟$ forgetting the map to $C$. (Strange to me, but it seems that forgetting here is the left adjoint.) Then we can apply the monad $RL$ to $B\to C$ to get a cobar resolution of $B$, take the product with $A$, and so get the cosimplicial object $A\times C\times B,A\times C\times C\times B,...$.

Perhaps this is the wrong section to discuss this, and when we figure it out, maybe we should make a section on fiber products of derived spaces.