Notes on Split Coequalizers

Jason Reed March 23, 2001

Let Sc be the category whose objects are 𝔄, 𝔅, 0 and whose arrows are

δ_{0}, δ_{1} : 𝔄 → 𝔅 Δ_{0}, Δ_{1}, id_{𝔄} : 𝔄 → 𝔄 σ : 𝔅 → 𝔄 id_{𝔅} : 𝔅 → 𝔅 {0_{M,N} : M → N | M, N ∈ {𝔄, 𝔅, 0}}

The composition law is given as follows:

δ_{0}σ = id_{𝔅} δ_{1}σ = 0_{𝔅𝔅} δ_{0}Δ_{i} = δ_{i} δ_{1}Δ_{i} = 0_{𝔄𝔅}

σδ_{i} = Δ_{i}

Δ_{0}Δ_{i} = Δ_{i} Δ_{1}Δ_{i} = 0_{𝔄𝔄} f0_{M,N} = 0_{M,cod f} 0_{M,N}f = 0_{dom f,N}

Sc sort of looks like this, without all the zeroes:

[figure: 𝔄; δ_{0}; δ_{1}; 𝔅; 𝔄; Δ_{0}; Δ_{1}; σ; 𝔅; ‖]

A fork is a diagram

[figure: A; d_{0}; d_{1}; B; e; C]

such that ed_{0} = ed_{1}. A split fork is a fork together with a splitting

[figure: A; s; B; t; C]

such that

[figure: B; s; A; d_{0}; B; C; e; t; B; d_{1}; e; C; e]

commutes, and the top two rows compose to identities.

Proposition 0.1 Sc is isomorphic to the category generated by the graph

[figure: G; :=; A; d_{0}; d_{1}; \underline{s}; B; e; \underline{t}; C]

with the identities ed_{0} = ed_{1}, d_{1}s = te, d_{0}s = id_{B}, and et = id_{C}.

Proof Let F: Grph → Cat be the left adjoint of the forgetful functor U: Cat → Grph. Let J = 2 + 2 + 2 + 2, and define f_{0}, f_{1} : J → F G by f_{0} := [ed_{0}, d_{1}s, d_{0}s, et] and f_{1} := [ed_{1}, te, id_{B}, id_{C}]. (That is, each arrow is defined by a copairing of arrows from 2 into F G, i.e. particular paths in G) Then Q, the category generated by G subject to the identities is by definition the coequalizer of f_{0}, f_{1}.

[figure: J; f_{0}; f_{1}; F G; q; Q]

To show Q ≅ Sc, it sufficies that Sc has the universal property of the coequalizer. We need a r: F G → Sc, and for every h: F G → H such that hf_{0} = hf_{1}, we want to show that there is a unique k such that

H

[figure: h]

[figure: F G; r; Sc; k]

commutes. Since F G is a free category, we only need to specify r’s action on G. The action on objects is obvious: we take A to 𝔄, B to 𝔅, C to 0. The action of arrows is equally apparent: we take d_{i} to δ_{i}, s to σ, e to 0_{𝔅,0} and t to 0_{0,𝔅}.

Now if we are to have kr = h, then we must have k(r(x)) = h(x) for all x ∈ {d_{0}, d_{1}, s, e, t}. That is,

k(δ_{i}) = h(d_{i})
k(σ) = h(s)
k(0_{𝔅,0}) = h(e)
k(0_{0,𝔅}) = h(t)

If k is to be a functor, then

k(Δ_{i}) = k(σδ_{i}) = k(σ)k(δ_{i}) = h(s)h(d_{i})
k(0_{0,𝔄}) = k(σ0_{0,𝔅}) = k(σ)k(0_{0,𝔅}) = h(s)h(t)
k(0_{𝔄,0}) = k(0_{𝔅,0}δ_{0}) = k(0_{𝔅,0})k(δ_{0}) = h(e)h(d_{0})
k(0_{𝔄,𝔅}) = k(0_{0,𝔅})k(0_{𝔄,0}) = h(t)h(e)h(d_{0})
k(0_{𝔅,𝔄}) = k(0_{0,𝔄})k(0_{𝔅,0}) = h(s)h(t)h(e)
k(0_{0,0}) = k(0_{𝔅,0})k(0_{0,𝔅}) = h(e)h(t)
k(0_{𝔄,𝔄}) = k(0_{0,𝔄})k(0_{𝔄,0}) = h(s)h(t)h(e)h(d_{0})
k(0_{𝔅,𝔅}) = k(0_{0,𝔅})k(0_{𝔅,0}) = h(t)h(e)

Using the fact that h coequalizes f_{0}, f_{1}, we can show that this indeed defines a functor.

So of course any split coequalizer is an arrow out of F G that coequalizes f_{0}, f_{1}, hence factors uniquely through r. So the split coequalizers in Sets, say, are the presheaves Sets^{Sc} = Sets^{(Sc^{op})^{op}}. The structure of the full subcategory of representable presheaves looks like the following: (4 stands for {0_{AA}, Δ_{1}, Δ_{0}, id_{A}} and 3 stands for {0_{AB}, δ_{1}, δ_{0}})

[figure: 4; 0122; 0001; 3; !; 1]

[figure: 2; 02; 00; = = = = = = = =; 00; 2; 02; 00; !; 1; ‖]

[figure: 1; 0; = = = = = = = = = = = = = = = =; 1; 0; = = = = = = = = =; 1; ‖]

[figure: 4; ⊃; 3; ⊃; 1]

[figure: 2; 0011; = = = = = = = = =; 2; 001; ⊃; 1; ‖]

[figure: 1; !; = = = = = = = = =; 1; !; = = = = = = = = =; 1; !]