Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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Filtered colimits commute with finite limits in Set

Statement

For every small filtered category J and finite category K, and every D:J×K→Set, the canonical comparison

colim⁡jlim⁡kD(j,k)⟶lim⁡kcolim⁡jD(j,k)

is a bijection. This includes the empty finite limit.

Facts & Assumptions

Given: The categories and diagram in the statement.

[F1]

Filteredness combines finitely many objects at a common later stage and coequalizes finitely many parallel arrows (Filtered categories and filtered colimits).

[L1]

Equality of two elements in a filtered Set-colimit occurs at one common later stage (Two representatives in a filtered colimit of sets are equal exactly when they become equal at one common later stage).

Proof

technique · representatives
1.1

By [L3], it suffices to prove that a filtered colimit preserves finite products and equalizers. For a finite product, a tuple on the right has finitely many coordinates, each represented at some stage. Repeated use of [F1] moves all representatives to one common stage, producing a tuple there. Thus the comparison is surjective.

F1L2L3
1.2

If two common-stage tuples have the same image, [L1] makes each coordinate equal at some later stage. There are finitely many coordinates, so repeated use of [F1] moves all those equalities to one stage. The tuples then agree, which proves injectivity.

F1L1L2
1.3

For an equalizer, an element on the right is represented by x at some stage and its two images become equal in the filtered colimit. By [L1], after moving x to one later stage its two images are equal there, so it is represented by a stagewise equalizer element. This proves surjectivity.

L1L2
1.4

If two stagewise equalizer elements become equal in the ambient filtered colimit, [L1] makes them equal at a common later stage; functoriality keeps them inside the later equalizer. This proves injectivity.

L1L2
1.5

For the empty product, each stage and the target are singletons. The filtered category is nonempty by [F1], so the colimit of the constant singleton diagram is a singleton, not empty.

F1L2
2.1

Steps 1.1 to 1.5 make every finite-limit comparison bijective. By [L4], filtered colimits preserve finite limits, proving the displayed assertion.

L3L4step 1.1step 1.2step 1.3step 1.4step 1.5∎

Depends on

Used by

Dependency tree · two levels

22 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