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There is a well-known way to conclude the cocompleteness of a category from its completeness. Namely, if a category is complete, well-powered and has a cogenerating set, then it is cocomplete (easy corollary of the special adjoint functor theorem).

I was hoping I could use this to prove that the category $\mathbf{Grp}$ of groups is cocomplete. I would find it interesting, for completeness of $\mathbf{Grp}$ is easy to prove (the forgetful functor to the category of sets creates limits), while cocompletenes is not so easy (constructing free product is quite tedious).

But alas, a quick google search revealed that $\mathbf{Grp}$ doesn't have a cogenerator (it can be read off the only freely available page of this article). I suspect it doesn't have a cogenerating set, either.

Is there a categorical proof of the cocompleteness of $\mathbf{Grp}$?

Edit: I'd like to point out that I found out afterwards the following exercise 1 in section IX.1 of Mac Lane's CWM (I didn't expect to find this exercise in the chapter on "Special Limits"!)

Use the adjoint functor theorem to prove in one step that Grp has all small colimits.

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$\text{Grp}$ doesn't have a cogenerating set because there are simple groups of arbitrarily large cardinality (see for example…). – Qiaochu Yuan Aug 9 '12 at 2:58
Why not use the general adjoint functor theorem? I think the cardinality trick works for $[\mathcal{J}, \textbf{Grp}]$. – Zhen Lin Aug 9 '12 at 4:48
Here is a general nonsense proof that the category of algebras for a monad on $\textbf{Set}$ is cocomplete. – Zhen Lin Aug 9 '12 at 5:03
@Zhen Lin, shouldn't you just put that nLab link into an answer? – Omar Antolín-Camarena Aug 9 '12 at 16:23
@Bruno: if $G_i$ is a set-indexed family of groups, find a simple group $H$ of cardinality larger than $\bigcup G_i$. Then there is no nonzero map from $H$ to any of the $G_i$, so $G_i$ is not a cogenerating set. – Qiaochu Yuan Aug 9 '12 at 19:09
up vote 6 down vote accepted

I think a proof using the general adjoint functor theorem is closest in spirit to a general nonsense proof, but it involves some rather concrete arguments about cardinality!

Let $\mathcal{J}$ be a small category, and let $\Delta : \textbf{Grp} \to [\mathcal{J}, \textbf{Grp}]$ be the "diagonal" functor. $\textbf{Grp}$ has colimits of shape $\mathcal{J}$ if and only if this functor $\Delta$ has a left adjoint (provided we assume a sufficiently large axiom of choice, or be careful about defining what we mean by "has colimits of shape $\mathcal{J}$"). Now recall the general adjoint functor theorem: when $\mathcal{D}$ is complete and locally small, a functor $U : \mathcal{D} \to \mathcal{C}$ has a left adjoint if and only if, for each object $X$ in $\mathcal{C}$, the comma category $(X \downarrow U)$ has a small weakly initial family (a.k.a. a solution set). So, for $U = \Delta$ and a diagram $X : \mathcal{J} \to \textbf{Grp}$, we must find a small set of groups such that every cocone on $X$ factors through one of those groups. But this is easy enough to arrange: let $\kappa$ be the cardinality of the disjoint union of all the groups $X_j$; then there is only a set of isomorphism classes of groups generated by $\kappa$ or fewer elements. (To be very concrete about it, we could say "the set of all groups whose underlying set is a subset of $\aleph_0 \cdot \kappa$".)

On the other hand, the above proof is morally appealing to the existence of coproducts in $\textbf{Grp}$, or at least free groups on arbitrarily many generators. So maybe it would be better to use a more direct argument. This nLab article sketches a proof that the category of algebras for a monad on $\textbf{Set}$ is always cocomplete; one then uses the fact that $\textbf{Grp}$ is monadic over $\textbf{Set}$ to conclude that $\textbf{Grp}$ is cocomplete.

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Thanks, Zhen Lin! I'm not familiar with algebras for a monad, but the first proof using GAFT was very useful. I'll post my own writeup of the proof with some more detail in a little while. – lentic catachresis Sep 4 '12 at 1:48

Zhen Lin already sketched a proof of the cocompleteness of $\mathbf{Grp}$ using the general adjoint functor theorem. Here's how I filled up the details.

Let $J$ be a small category; we wish to show the diagonal functor $\Delta:\mathbf{Grp}\to \mathbf{Grp}^J$ has a left adjoint. The diagonal functor preserves limits (direct consequence of the pointwise construction of limits in $\mathbf{Grp}^J$) and $\mathbf{Grp}$ is complete, so it suffices to check the solution set condition.

Let $D\in \mathbf{Grp}^J$ be a diagram. We note $D_j:=Dj$ for all $j\in J$.

Consider the class consisting of groups of cardinality $\aleph_0$ and of groups of cardinality $\leq \lvert \bigsqcup\limits_{j\in J} D_j \rvert$ for all $j\in J$. We consider the set $\{G_\alpha\}_{\alpha\in \Lambda}$ of groups consisting of one isomorphism class per each of these groups.

I claim the set of all natural transformations $\{D\Rightarrow \Delta G_\alpha\}_{\alpha\in \Lambda}$ is a solution set.

Fix $\varphi:D\Rightarrow \Delta G$. We must check that there exists $\alpha\in \Lambda$, $\tau^\alpha:D\Rightarrow \Delta G_\alpha$ in the solution set and $t:G_\alpha\to G$ a group homomorphism such that the following diagram commutes:

enter image description here

Remember that natural transformations $D\Rightarrow \Delta K$ are exactly cocones for the diagram $D$ with vertex $K$.

So, to put it differently, we must find a cocone morphism between some cocone $\tau^\alpha$ in the solution set and the fixed cocone $\varphi$, i.e. a group morphism $t:G\alpha\to G$ such that the following diagram commutes for every arrow $e^i_j:i\to j$ of $J$:

enter image description here

Let $H$ be the subgroup of $G$ generated by $\bigsqcup\limits_{i\in J} \operatorname{Im} \varphi_j$. Then $H=G_\alpha$ for some $\alpha\in \Lambda$ (by the same argument used in the proof that there exist free groups using GAFT).

We define $\tau^\alpha_i:D_i\to G_\alpha$ as $\tau^\alpha_i(x)=(\varphi_i(x),i)$ for every $i\in J$. So $\tau^\alpha:D\Rightarrow \Delta G_\alpha$ is a cocone for $D$: this is quickly verified using the fact that $\varphi$ is already a cocone.

Now we define $t:G_\alpha\to G$ as $t(\varphi_i(x),i)=\varphi_i(x)$; clearly $t\circ \tau^\alpha_i=\varphi_i$ for all $i\in J$, thus finishing the proof.

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