Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 23 / 5 / Solution

A filtered category is nonempty, every pair of objects maps to a common object, and every parallel pair becomes equal after postcomposition. Equivalently every finite diagram in a category has a cocone.

For a filtered set diagram $D:I\to \mathbf{Set}$, the finite-stage equality in a filtered set colimit says that representatives $x\in D(i)$ and $y\in D(j)$ define the same element exactly when their images agree at some common stage. Indeed, equality in the ordinary disjoint-union-and-quotient construction is witnessed by a finite zigzag; filteredness moves its objects to a common stage and equalizes the finitely many competing arrows.

For a finite categorical product, choose a common stage for all coordinate representatives to obtain surjectivity of the colimit-to-product comparison. Finitely many coordinate equalities can then be realized together at one later stage, proving injectivity. The empty product is preserved because a nonempty filtered category is connected. For an equalizer, an element whose two images agree in the colimit reaches a stage where they agree, so it is represented by an element of the stagewise equalizer; the same equality criterion proves injectivity. Finite products and equalizers construct every finite limit. Hence filtered colimits commute with finite limits in sets.

We prove the equivalence by the cycle solution, solution, solution, solution, then back to solution.