When is the kernel pair of a finite presentation of algebraic structures finitely generated?

Question

Fix an algebraic theory; denote its free models by $T^k$.

There are two possible definitions of what it means for a coequalizer $T^m\twoheadrightarrow M$ to be a finite presentation of $M$.

1. $f\colon T^m\twoheadrightarrow M$ is the coequalizer of some pair of morphisms $T^k\rightrightarrows T^m\twoheadrightarrow M$. This is the standard definition in category theory textbooks.
2. If $K$ is a kernel pair $K\rightrightarrows T^m\twoheadrightarrow M$ of $f\colon T^m\twoheadrightarrow M$, then $K$ is finitely generated in the sense that there is some coequalizer $T^k\twoheadrightarrow K$. This is the standard definition in algebra textbooks.

It is easy to see that the second notion implies the first: $T^m\twoheadrightarrow M$ is the coequalizer of the composites $T^k\twoheadrightarrow K\rightrightarrows T^m$. Conversely, the notions are equivalent in the standard examples (groups, $R$-modules, rings). Are the notions always equivalent, or do we need certain conditions on the algebraic theory (e.g. being Mal'cev)?

(Note: the second notion of $M$ being finitely presented is better behaved in the sense that it implies that every coequalizer $T^{m'}\twoheadrightarrow M$ has a finitely generated kernel.)

Answer 1

The sense in which the kernel pair is finitely generated specified by the question is incorrect. What is actually happening is this.

Given a coequalizer $X\rightrightarrows Y\to Z$, we have equalizers $\mathcal C(X,F(i))\leftleftarrows\mathcal C(Y,F(i)\hookleftarrow\mathcal C(Z,F(i)$ for each object $F(i)$ of a filtered diagram $F$ (because representable functors preserve limits). Since filtered colimits commute with equalizers, we get an equalizer $\varinjlim\mathcal C(X,F(i))\leftleftarrows\varinjlim\mathcal C(Y,F(i))\hookleftarrow\mathcal\varinjlim C(Z,F(i))$ with comparison maps from the equalizer $\mathcal C(X,\varinjlim F(i))\leftleftarrows\mathcal C(Y,\varinjlim F(i))\hookleftarrow\mathcal C(Z,\varinjlim F(i))$.

By https://math.stackexchange.com/a/4408111/400, respectively by definition, a component of the comparison maps is, for all filtered diagrams $F$, a monomorphism, respectively isomorphism, if and only if the corresponding object is finitely generated, respectively finitely presentable.

If $X$ is finitely generated and $Y$ is finitely presentable, then the top parallel pair factors as a monomorphism followed by the bottom parallel pair, whence they have isomorphic equalizers, i.e. $Z$ is finitely presentable. It is in this sense that a relation defines a finitely presentable object if it's finitely generated: $Y\to Z$ is the coequalizer of $X\to K\rightrightarrows Y$ where $K\rightrightarrows Y$ is the kernel pair and $X$ is finitely generated.

Conversely, if $X$ is a filtered colimit of finitely presentable objects, then $Y$ and $Z$ being finitely presentable imply $Y\to Z$ is also the coequalizer of $W\rightrightarrows Y\to Z$ for a finitely presentable $W$ (with a natural map to $X$) (see https://mathoverflow.net/a/382672/75650). Applying this to $X\rightrightarrows Y$ the kernel pair of $Y\to Z$ shows that any relation defining a finitely presentable object is finitely generated in the above sense.

Note that the kernel pair does not have to be finitely generated. Indeed, we have a cokernel $\mathbb Z\to F_2\to\mathbb Z^2$ given by $1\mapsto xyx^{-1}y^{-1}$, $x,y\mapsto(1,0),(0,1)$ showing $\mathbb Z^2$ is finitely presentable, but its kernel, the derived subgroup of $F_2$, is not finitely generated. It should also be the case that the kernel pair, which is the semi-direct product of $F_2$ and its derived subgroup, is also not finitely generated.