nLab wide pullback (Rev #15)

Contents

Definition

A fibre product or fiber product is a product in a slice category π’žβ†“C\mathcal{C} \downarrow C. The fibre product of two morphisms f:Aβ†’Cf: A \to C, g:Bβ†’Cg: B \to C is the same as their pullback; accordingly, a fiber product more than two morphisms f i:A iβ†’Cf_i : A_i \to C is often called a wide pullback.

More explicitly, for f:Aβ†’Cf : A \to C and g:Bβ†’Cg : B \to C two morphisms in a category π’ž\mathcal{C}, the fiber product AΓ— CBA \times_C B of AA with BB over CC is, if it exists, the pullback

AΓ— CB β†’ B ↓ ↓ g A β†’f C. \array{ A \times_C B &\to& B \\ \downarrow && \downarrow^g \\ A &\stackrel{f}{\to}& C } \,.

This term comes from thinking of AA and BB as bundles over CC; then the fiber of AΓ— CBA \times_C B over a generalized element xx of CC is the product of the fibers of AA and BB over xx. In other words, the fiber product is the product taken fiber-wise.

Of course, the fiber of AA at the generalized element x:I→Cx: I \to C is itself a pullback I× CAI \times_C A; the terminology depends on your point of view.

You're English, David, so how come you don't like β€˜fibre’? β€”Toby

I just thought it was policy here. That’s what the HowTo page says anyway. β€”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. β€”Toby

Examples

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

AΓ— CB={(a,b)∈AΓ—B|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 AΓ—B,AΓ—CΓ—B,AΓ—CΓ—CΓ—A,...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

AΓ—Bβ†’(Id A∘p 1)Γ—(f∘p 1)Γ—(Id B∘p 2)AΓ—CΓ—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

AΓ—Bβ†’(Id A∘p 1)Γ—(g∘p 2)Γ—(Id B∘p 2)AΓ—CΓ—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 AΓ— CBA \times_C B is the equalizer of these two maps, i.e. the limit over the diagram

A×B→→A×C×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 β€œinner” ones come from the diagonal Cβ†’CΓ—CC \to C \times C and the outer ones from ff and gg as above. So for instance the next three would be

AΓ—CΓ—Bβ†’(Id A∘p 1)Γ—(f∘p 1)Γ—(Id C∘p 2)Γ—(Id B∘p 3)AΓ—CΓ—CΓ—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

AΓ—CΓ—Bβ†’(Id A∘p 1)Γ—(Id CΓ—Id C∘p 2)Γ—(Id B∘p 3)AΓ—CΓ—CΓ—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

AΓ—CΓ—Bβ†’(Id A∘p 1)Γ—(Id C∘p 2)Γ—(g∘p 3)Γ—(Id B∘p 3)AΓ—CΓ—CΓ—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×B→→A×C×B→→→A×C×C×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 AΓ— CBA \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 AΓ— CBA \times_C B in a homotopically meaningful way, we should somehow resolve BB. So if π’Ÿ\mathcal{D} is our category and π’Ÿ C\mathcal{D}_C is the category of objects over Cβˆˆπ’ŸC \in \mathcal{D}, then we have an adjoint pair of functors R:=CΓ—?:π’Ÿβ†’π’Ÿ CR:= C\times? : \mathcal{D} \rightarrow \mathcal{D}_C and L:=π’Ÿ Cβ†’π’ŸL:=\mathcal{D}_C \rightarrow \mathcal{D} forgetting the map to CC. (Strange to me, but it seems that forgetting here is the left adjoint.) Then we can apply the monad RLR L to Bβ†’CB \rightarrow C to get a cobar resolution of BB, take the product with AA, and so get the cosimplicial object AΓ—CΓ—B,AΓ—CΓ—CΓ—B,...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.

Revision on July 4, 2010 at 13:45:44 by Todd Trimble See the history of this page for a list of all contributions to it.