Coequalizer that is not absolute

Question

A coequalizer is called absolute when it is preserved by each functor. Could somebody give me an example of a coequalizer that is not absolute (with proof) ? If possible it would be great if the example were in Set.

Answer 1

I have one, I hope it is what you were looking for. A particular type of coequalizer in an additive category is the cokernel defined by $\operatorname{coker}(f) := \operatorname{coeq}(f,0)$.

Let us consider the abelian category Ab of abelian groups (or equivalently $\mathbb{Z}$-Mod).

We know that an additive (preserves $\oplus$) functor is right-exact iff it sends right exact sequences to right exact sequences or equivalently if it preserves any colimit... We just need to find one "bad" functor, I don't want to think too much, so I will take a left exact-functor which seems to be bad enough: $F := \hom(-,\mathbb{Z})$.

One quick remark is that we have a short exact sequence $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$ where the projection is the cokernel of the multiplication by two. But since $\hom(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}) = 0$, $F$ sends this sequence (up to isomorphism) to $$\mathbb{Z} \to \mathbb{Z} \to 0$$ where the first morphism is the multiplication by $2$, so we get $$F(\operatorname{coker} (2\cdot -)) = 0 \neq \mathbb{Z}/2\mathbb{Z} = \operatorname{coker} (F( 2 \cdot -)) $$

Which means that $\mathbb{Z}/2\mathbb{Z}$ is not absolute

Answer 2

My question was answered by the comment from Geoffrey Trang. The two maps from $\{1\}$ to $\{1,2\}$ have a coequalizer that is not preserved by the squaring endofunctor $X \mapsto X^2$ of Set.