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The category $\mathbf{FreeAb}$ of free abelian groups and group homomorphisms is a nice example for the dichotomy of nice objects and nice categories. Its objects are very nice, but the category is badly behaved. This question is about giving a proof that it is badly behaved.

Question. What are some categorical properties that $\mathbf{FreeAb}$ does or does not satisfy, and how to prove that? Specifically, does the category $\mathbf{FreeAb}$ have filtered colimits? Is it well-copowered?

Caution. When dealing with these questions one has to be very careful. For example, $\mathbf{FreeAb}$ does not have cokernels by MSE/765730, but the fact that $\mathbf{FreeAb} \subseteq \mathbf{Ab}$ is not closed under cokernels (which is trivial) does not prove anything. A potential cokernel could look differently, i.e., it does not have to be preserved by the inclusion functor to $\mathbf{Ab}$. See also the discussion about epimorphisms below.

Here is my progress.

The category $\mathbf{FreeAb}$ is additive, has all coproducts, and it has equalizers (use this). Hence, it has pullbacks. As mentioned, it doesn't have coequalizers. The forgetful functor $\mathbf{FreeAb} \to \mathbf{Set}$ is representable (by the integers), hence preserves all limits. In particular, the monomorphisms are the injective homomorphisms, from which it follows immediately that the category is well-powered. One can also deduce that it doesn't have products (again, this is much more than just saying that $\mathbf{FreeAb} \subseteq \mathbf{Ab}$ is not closed under products, which is witnessed by $\mathbb{Z}^{\mathbb{N}}$ as we all know).

However, epimorphisms do not coincide with surjective homomorphisms. For example, $2 : \mathbb{Z} \to \mathbb{Z}$ is an epimorphism in the category of torsionfree abelian groups, hence also in $\mathbf{FreeAb}$. This shows that $\mathbf{FreeAb}$ is not balanced. In general, one can show that $f : A \to B$ is an epimorphism iff $f(A)$ is not contained in a proper direct summand of $B$. For this reason, I do not know if the category is well-copowered (but I think it is).

Filtered colimits can also look different. For example, the colimit of $$\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{2} \cdots$$ actually does exist, it is $0$. This is because in a free abelian group every element which is divisible by $2^n$ for all $n$ must be zero. The colimit in $\mathbf{Ab}$ is $\mathbb{Z}[1/2]$, this is different. But I suspect that filtered colimits don't have to exist in general. Some filtered colimits are actually preserved, for example $\mathbb{Z} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Z}^2 \hookrightarrow \dotsc$ with transition maps $x \mapsto (\dotsc x,0)$, which has colimit $\mathbb{Z}^{\oplus \mathbb{N}}$ in both categories.

Martin Brandenburg
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1 Answers1

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It does not have filtered colimits, but it is well-copowered.

Filtered colimits

If we have some diagram in $\mathbf{FreeAb}$ that has a colimit $\hat{C}$ in $\mathbf{FreeAb}$ and has colimit $C$ in $\mathbf{Ab}$, then the natural map $C\to\hat{C}$ (given by the universal property of $C$) is the universal map from $C$ to a free abelian group. $\hat{C}$ is the liberation of $C$, one might say.

Every torsion free abelian group is the filtered colimit of its finitely generated free abelian subgroups, but not every torsion free abelian group has a "liberation"; for example, the Baer-Specker group $\mathbb{Z}^{\mathbb{N}}$ does not, by this answer.

So in $\mathbf{FreeAb}$ there is no colimit of the filtered diagram of finitely generated free abelian subgroups of $\mathbb{Z}^{\mathbb{N}}$.

Well-copowered

Let $\alpha:\mathbf{Z}^{(I)}\to \mathbf{Z}^{(J)}$ be a map of free abelian groups. Then $\operatorname{im}\alpha$ is supported on $K$ for some $K\subseteq J$ of cardinality at most $\max\{\aleph_{0},\vert I\vert\}$. So if $\vert J\vert>\max\{\aleph_{0},\vert I\vert\}$, then $\operatorname{im}\alpha$ is contained in a proper summand of $\mathbf{Z}^{J}$, and so can't be an epimorphism.

So, given a free abelian group $A$, there is a bound on the cardinalities of the $B$ that can occur in an epimorphism $A\twoheadrightarrow B$. So $\mathbf{FreeAb}$ is well-copowered.

Jeremy Rickard
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