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

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.

Assume $F$ preserves finite limits. Given any finite diagram in the comma category $(A\downarrow F)$, first take its underlying limit $c$ in $\mathcal{C}$. Its structure maps $A\to F{c}_j$ form a cone, and preservation gives a unique map $A\to Fc$. This object and the underlying projections satisfy the limiting universal property in the comma category. The construction also includes the empty diagram, since $F$ sends a terminal object to a singleton. Thus $(A\downarrow F)$ has finite limits for every set $A$, proving (ii).

Take $A=1$. The category of elements $\int F=(1\downarrow F)$ has finite limits. Reversing its limiting cones gives cocones over every finite diagram in $(\int F{)}^{\mathrm{op}}$; a terminal object before reversal ensures nonemptiness. Its opposite category is therefore a filtered category, proving (iii).

Use the covariant density presentation. An object of $\int F$ is $(c,x)$ with $x\in Fc$; an arrow $(c,x)\to (d,y)$ is $u:c\to d$ with $F(u)x=y$. For each such object the Yoneda lemma gives a natural transformation $\mathcal{C}(c,-)\to F$. An arrow $u$ induces $\mathcal{C}(d,-)\to \mathcal{C}(c,-)$ by precomposition, so these transformations form a cocone indexed by $(\int F{)}^{\mathrm{op}}$.

At $a$, send a representative $(c,x,v:c\to a)$ to $F(v)x$. Every $z\in Fa$ comes from $(a,z,1_a)$. If two representatives have the same value $z$, their arrows to $(a,z)$ in $\int F$ identify both with that same representative in the opposite-indexed colimit. Thus the map is bijective and natural, provingThe index is small because $\mathcal{C}$ is small and every $Fc$ is a set. By (iii) it is a filtered category. This proves (iv).

Every covariant representable functor $\mathcal{C}(c,-)$ preserves categorical limits, since a morphism out of $c$ into a limit is precisely a compatible cone. A filtered colimit of these functors is computed pointwise, and filtered colimits commute with finite limits in sets. Interchanging the finite limit and filtered colimit therefore proves that $F$ preserves finite limits, which is (i). This completes the cycle. It is also the flat functor and finite-limit preservation criterion for a finitely complete small category.