Question

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?

Answer 1

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.

Answer 2

I think this is true. Let $G= \langle g_1, \ldots, g_n; R_1, \ldots, R_m \rangle$ be a finite presentation for $G$. Let $p:G\to G$ be a projection (or retract). Then $p(G)< G$ is generated by $p(g_1), \ldots, p(g_n)$. Let $F_n$ be the free group on $n$ generators, and $R\in F_n$ a relator of $p(G)$, so $R(p(g_1), \ldots, p(g_n))=1$. I claim that $p(R_1),\ldots, p(R_m)$ give finitely many relations for $p(G)$. Give a new presentation for $G$, $$\langle g_1, \ldots, g_n , y_1, \ldots, y_n ; y_1^{-1} p(g_1) , \ldots, y_n^{-1} p(g_n), R_1, \ldots, R_m \rangle.$$ With respect to this presentation, the homomorphism $p$ is given by $p(g_i)=p(g_i), p(y_i)=p(g_i)$. Then we must have the relation $R(y_1, \ldots, y_n)=1 \in G$, and therefore $R(y_1, \ldots, y_n) $ may be written as a product of conjugates of the relators. Then $p(R(y_1,\ldots,y_n)) = R(p(g_1),\ldots,p(g_n))$ is a product of conjugates of projections of the relators of the second presentation of $G$. This shows that $p(G)$ is finitely presented, and since the first $n$ relators project to the trivial relator, we see that the relators are given by $p(R_1),\ldots, p(R_m)$.

Here's a reformulation:

Let $G$ be a finitely presented group, with a projection $p:G\to G$, so $p^2=p$.

Then there is a free group $F = \langle g_1, \ldots, g_n\rangle$ and an epimorphism $\phi:F\to G$ such that $ker(\phi)$ is finitely normally generated. We may write the presentation as $$ \langle g_1,\ldots, g_n; R_1,\ldots, R_m\rangle.$$ The kernel of $p$ is normally generated by $\phi(g_i) p(\phi(g_i))^{-1}$ (exercise)*. Let $p': F\to F$ be a lift of $p$, which exists by the universal property of $F$, so that $p\circ \phi = \phi \circ p'$. Then $ker(p\circ \phi) = \phi^{-1}( ker(p)) $ is normally generated by elements projecting to the normal generators of $ker(p)$ togther with $ker(\phi)$. We may take $g_i p'(g_i)^{-1}$ together with the finitely many normal generators of $ker(\phi)$ to get normal generators of $ker(p\circ \phi)$. This gives a finite presentation of $p(G)$, given by $$\langle g_1,\ldots,g_n ; R_1,\ldots, R_m, g_1 p'(g_1)^{-1},\ldots, g_n p'(g_n)^{-1} \rangle.$$

In my previous "argument", I attempted to project this presentation by $p'$ to $p'(F)$, which is a free subgroup of $F$, but this is unnecessary, and I was incorrect that the new relators project to trivial relators in $p'(F)$.

*(exercise) Here's the details to show the kernel is normally generated by $\phi(g_i) p(\phi(g_i))^{-1}$. For simplicity of notation, I'll denote the generators as $g_i$ instead of $\phi(g_i)$, and assume that they are semigroup-generators of $G$. Then $ker(p)$ is normally generated by $g_i p(g_i)^{-1}$. Call this normal subgroup $K$. Notice that it suffices to show that $g p(g)^{-1} \in K$, for all $g\in G$, since if $g\in ker(p)$, then $g p(g)^{-1} = g \in K$. We may show this by induction on the length of the word representing $g$. Clearly it's true if $g= g_i$. Now, suppose it is true for products of $k$ generators, $k\in \mathbb{N}$. Then if $g= g_{i_1} \cdots g_{i_{k+1}}$, we have $$g p(g)^{-1}= g_{i_1} \cdots g_{i_{k+1}} p(g_{i_{k+1}})^{-1} \cdots p(g_{1})^{-1} = g_{i_1} p(g_{i_1})^{-1} ( p(g_{i_1}) ( g_{i_2} \cdots g_{i_{k+1}} p(g_{i_{k+1}})^{-1} \cdots p(g_{i_2})^{-1} ) p(g_{i_1})^{-1}) \in K$$ by induction and the fact that $K$ is closed under products and conjugations.