let's call an object $x$ of a cocomplete category (categorical) finitely generated if $\hom(x,-)$ commutes with filtered colimits of monomorphisms, and finitely presented if $\hom(x,-)$ even commutes with arbitrary filtered colimits. I know that in the literature there are some other definitions for these notions. at least in categories of algebraic structures, it would be nice that these notations coincide with the usual ones. let's concentrate on the category of groups, for these make problems ;).


it's easy to see that categorical finitely generated means the usual finitely generated. but I've figured out that a group $G$ is categorical finitely presented if and only if there is a finitely presented group $H$, over which $id\_G : G \to G$ factors through. is this the same as beeing finitely presented?


the question is equivalent to: let $H$ be a finitely presented group and $p : H \to H$ a projection, i.e. a homomorphism satisfying $p^2 = p$. is then $im(p)$ also finitely presented?



projections of finitely presented groups
Here is a direct proof that "categorically finitely presented" (which I think it is more usual to call "finitely presentable," since no actual presentation is necessarily given) is the same as "finitely presented" in the usual sense. This is a special case of Theorem 3.12 in Adamek and Rosicky's book "Locally presentable and accessible categories," which applies to all finitary varieties of algebras.


First suppose $G$ is finitely presented in the usual sense, with $X$ a finite set of generators and $R \subset F[X]$ a finite set of relations, where $F[-]$ is the free group functor. Then there is a coequalizer diagram $F[R] \rightrightarrows F[X] \twoheadrightarrow G$. Since finite colimits of c.f.p. objects are c.f.p. (since finite limits commute with filtered colimits in Set), it suffices to show that finitely generated *free* groups are c.f.p. But $Hom\_{Grp}(F[X],H) \cong Hom\_{Set}(X,U(H))$ where $U$ is the underlying-set functor, which preserves filtered colimits, and finite sets are c.f.p. in Set (this is easy to see).


Conversely, suppose $G$ is c.f.p. Then in particular it is c.f.g. and hence (as you say) finitely generated in the usual sense. One way to see that is to write $G$ as the directed union (= filtered colimit of monics) of its finitely generated subgroups; then since $G$ is c.f.g. its identity must factor through one of these subgroups, and hence that subgroup must be all of $G$. Now let $X$ be a finite set of generators and consider the diagram of groups whose vertices are all finitely presented quotients $F[X]\twoheadrightarrow H\_i$ through which $F[X]\twoheadrightarrow G$ factors, and whose transition maps $k\_{i,j}\colon H\_i \to H\_j$ are maps under $F[X]$ (and hence over $G$). Then the canonical maps $h\_i\colon H\_i \to G$ make $G$ into the colimit of this diagram, since any relation holding in $G$ must hold in some finitely presented quotient of $F[X]$. Moreover, this diagram is filtered, so $id\_G\colon G\to G$ factors through some $H\_i$ via a map $u\colon G \to H\_i$ with $h\_i u = id\_G$.


I'm guessing this is the point to which you got. The clever argument from A&R then goes as follows. By the first part of the argument, $H\_i$ is c.f.p. But we have two maps $id\_{H\_i}\colon H\_i\to H\_i$ and $u h\_i\colon H\_i\to H\_i$ which become identified in the colimit, i.e. $h\_i id\_{H\_i} = h\_i u h\_i$; thus there must be some $k\_{i,j}\colon H\_i \to H\_j$ such that $k\_{i,j} = k\_{i,j} u h\_i$. It then follows that $h\_j\colon H\_j \to G$ is an isomorphism with inverse $k\_{i,j} u$, for on the one hand, $h\_j k\_{i,j} u = h\_i u = id\_G$, but on the other hand, $k\_{i,j} u h\_j k\_{i,j} = k\_{i,j} u h\_i = k\_{i,j}$ and $k\_{i,j}$ is epic. Thus, $G$ is finitely presented since $H\_j$ is so.


