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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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Every small limit can be constructed as an equalizer between products over the objects and arrows of the index category

Statement

Let D:J→C be small. If the products

P=∏j∈Ob⁡JD(j),Q=∏u:j→kD(k)

and the equalizer of the maps s,t:P⇉Q defined by qus=D(u)pj and qut=pk exist, then that equalizer is a limit of D.

Facts & Assumptions

Given: The displayed products and an equalizer e:L→P of s,t.

[F2]

An equalizer represents arrows on which its parallel pair agrees (Equalizers and coequalizers as limits and colimits of a parallel pair).

[F3]

Proof

technique · construction
1.1

By [F3] the two displayed families are set-indexed. By [F1], the stated coordinate equations define unique maps s,t:P→Q.

F1F3
2.1

Put λj=pje. Since se=te, equality of the u-coordinates says D(u)λj=λk for every u:j→k. Thus λ is a cone.

F2step 1.1
2.2

Given a cone ξj:X→D(j), [F1] supplies a unique a:X→P with pja=ξj. Its cone equations imply qusa=D(u)ξj=ξk=quta for all u, so product uniqueness gives sa=ta.

F1givenstep 1.1
3.1

By [F2], a factors uniquely as a=eh with h:X→L. Then λjh=pjeh=ξj, so h is a cone morphism.

F2step 2.1step 2.2
4.1

If h′ has the same leg equations, product uniqueness gives eh′=a=eh; equalizer uniqueness gives h′=h.

F1F2step 3.1
5.1

Steps 2.1, 2.2, 3.1, and 4.1 are exactly [F4], so (L,λ) is a limit. If J is empty, both products are terminal objects, s=t, and their equalizer is isomorphic to the terminal object, so the construction still applies.

F1F2F4step 2.1step 3.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources