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Under dependent choice, algebras for a finitary monad on a complete cocomplete locally small category have coequalizers

Statement

Assume dependent choice. If T is a finitary monad on a complete, cocomplete, locally small category C, then the Eilenberg–Moore category CT has coequalizers.

Facts & Assumptions

Given: Dependent choice, a complete cocomplete locally small category C, a finitary monad T, and algebra homomorphisms f,g:(A,a)⇉(B,b).

[L1]

A functor is finitary when it preserves every small filtered colimit (Finitary functors and finitary monads).

[L2]

Dependent choice produces a sequence from a nonempty set with an entire successor relation and a specified starting point (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain).

[L3]

Let V:A→B have complete locally small domain and be continuous. If V satisfies the solution-set condition at B∈B, then (B↓V) has an initial object, equivalently a universal arrow from B to V (General adjoint functor theorem, objectwise initial-object form).

[L4]

The Eilenberg–Moore forgetful functor strictly creates every limit existing in the base (The Eilenberg–Moore forgetful functor strictly creates every limit that exists in the base).

[L5]

If A and the indexing category J are small and chosen limits of the pointwise diagrams exist, those choices form a limit in [A,C] (For small source and index categories, chosen target limits and colimits compute the corresponding functor-category limits and colimits pointwise).

Proof

technique · direct
1.1givenconstruct

In C, form q0:B→Q0, the coequalizer of f,g, and p0:TB→P0, the coequalizer of Tf,Tg. Their universal properties induce maps u0:P0→Q0 and v0:P0→TQ0 characterized by u0p0=q0b and v0p0=Tq0. Both p0 and q0 are epimorphisms.

1.2L4L5construct

The category CT is complete by [L4] and locally small because its hom-sets are subsets of those of C. The walking parallel-pair category is finite and hence small. For any small limit diagram, the finitely many pointwise limits can be chosen without a choice axiom, so [L5] makes the parallel-pair functor category complete. The constant-diagram functor is continuous because these limits are pointwise.

2.1step 1.1algebraconstruct

Suppose (Pn,Qn,un,vn,pn,qn) has been constructed with unpn=qnb and vnpn=Tqn. Put Pn+1=TQn, and choose un+1:TQn→Qn+1 as a coequalizer of T(un),μQnT(vn):TPn⇉TQn. Define qn,n+1=un+1ηQn, vn+1=Tqn,n+1, pn+1=vnpn, and qn+1=qn,n+1qn. Naturality of η and the monad unit law give qn,n+1un=un+1ηQnun=un+1vn. Consequently un+1pn+1=qn+1b and vn+1pn+1=Tqn+1.

2.2step 1.1constructalgebra

Let h:(B,b)→(C,c) be any algebra homomorphism with hf=hg. Factor it uniquely as h=k0q0. Precomposing with the epimorphism p0 and using step 1.1 gives cT(k0)v0=k0u0.

3.1step 1.1step 2.1L2choose

To justify the countable sequence of choices in step 2.1, encode each finite stage by a set. For a state x, let S(x) be all valid successor codes of least possible von Neumann rank; it is a nonempty subset of some Vα, hence a set. Starting with the code from step 1.1, recursively close under x↦S(x) for finitely many steps and take the union over N. Replacement and union make this closure a set on which the successor relation is entire. Applying [L2] to that set gives one compatible sequence of stages.

3.2step 2.1step 2.2algebraconstruct

Suppose kn:Qn→C satisfies cT(kn)vn=knun. The algebra law for c shows that cT(kn) coequalizes T(un) and μQnT(vn), so it factors uniquely through un+1 as a map kn+1:Qn+1→C with kn+1un+1=cT(kn). The definitions in step 2.1 then give kn+1qn,n+1=kn and cT(kn+1)vn+1=kn+1un+1, closing the induction.

4.1step 2.1step 3.1construct

Form the sequential colimits Qω=colim⁡nQn and Pω=colim⁡nPn, with injections qn,ω and pn,ω. The identities qn,n+1un=un+1vn make the un a map of the two sequential diagrams, hence induce uω:Pω→Qω. The compatible qn induce qω:B→Qω. The shifted identities Pn+1=TQn and vn+1=Tqn,n+1 identify Pω with colim⁡nTQn.

5.1step 4.1L1

The natural-number indexing category is filtered, so finitarity [L1] identifies the colimit in step 4.1 with TQω. Make this identification, so Pω=TQω and the induced comparison vω:Pω→TQω is the identity. Thus uω:TQω→Qω.

6.1step 2.1step 5.1algebra

For every n, naturality of η and the definition of qn,n+1 give uωηQωqn,ω=qn,ω. The colimit injections are jointly epimorphic, so uωηQω=1Qω.

6.2step 5.1step 3.2

The compatible kn induce k:Qω→C. Passing the equations kn+1un+1=cT(kn) to the colimit gives kuω=cT(k), and compatibility at stage zero gives kqω=h.

7.1step 2.1step 5.1step 6.1algebra

The successor coequalizer equations are un+1T(un)=un+1μQnT(vn). Passing these compatible equations to the filtered colimit gives uωT(uω)=uωμQωT(vω)=uωμQω because vω=1 in step 5.1. Together with step 6.1, this makes (Qω,uω) a T-algebra.

8.1step 2.1step 5.1step 7.1

Passing unpn=qnb and vnpn=Tqn to the colimit gives uωTqω=qωb, so qω is an algebra homomorphism; it coequalizes f,g because q0 does.

9.1step 8.1step 6.2step 7.1construct

By steps 6.2 and 7.1, k is an algebra homomorphism, and step 6.2 gives kqω=h. Therefore the single algebra fork qω is a solution set for the constant-diagram functor Δ:CT→(CT)∙⇉∙ at the given parallel pair: every algebra fork factors through it, without a uniqueness assertion.

10.1step 9.1step 1.2L3∎

Apply [L3] to Δ using the singleton solution set from step 9.1 and the complete, locally small, and continuity properties proved in step 1.2. The resulting universal arrow is a left adjoint value for Δ, hence a coequalizer of f,g in CT. Since the pair was arbitrary, all coequalizers exist.

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