Absolute coequalizers in $\mathbf {Set} $

Question

Let $ A $ be a set and let $ R\subseteq A\times A $ be an equivalence relation on $ A $. Denote by $ p, q $ the projections $ R\longrightarrow A $ on the first and second factor, respectively. The coequalizer of $ p $ and $ q $ is then given by the projection $ A\longrightarrow A/R $ assigning to each element of $ A $ its equivalence class modulo $ R $.

Using the fact that epimorphisms in $\mathbf {Set} $ are split, or equivalently, the axiom of choice, it is shown in Handbook of Categorical Algebra, Vol. 1 by Francis Borceux, that this coequalizer is an absolute colimit, i.e. that it is preserved by any functor (see Proposition 2.10.2 ).

I was wondering:

Can we prove coequalizers of this kind absolute without using the axiom of choice?

Intuitively, it seems to me that we do not need any choices, but I don't know enough set theory to convince myself.

Answer 1

In fact, the following are equivalent (in ZF):

• The axiom of choice.
• Every coequaliser in $\mathbf{Set}$ is absolute.

Indeed, one can prove (in ZF) that every surjection is the coequaliser of its kernel pair, so it suffices to prove the following assertion:

The coequaliser diagram $$R \rightrightarrows A \rightarrow A / R$$ is absolute if and only $A \rightarrow A / R$ splits.

You have seen that a splitting of $A \to A / R$ makes the coequaliser diagram absolute, so I'll just prove the converse. By absoluteness, the functor $\mathrm{Hom}(A / R, -)$ preserves the coequaliser, and regular epimorphisms are surjective, so there must exist a morphism $A / R \to A$ whose composite with $A \to A / R$ is $\mathrm{id}_{A / R}$, which is what I claimed.