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The category of abelian groups $\mathsf{Ab}$ is the $\mathcal{Ind}$-completion of the full subcategory of finitely presentable abelian groups $\mathsf{Ab}_{fp}$. This is not so special, since the analogous statement holds for any finitary variety e.g. groups, rings, boolean algebras etc.

However one nice property of finitely presentable abelian groups is that they are closed under finite limits, in fact under finite products and subalgebras. Therefore $(\mathsf{Ab}_{fp})^{op}$ has finite colimits, which implies that it forms (essentially) the finitely presentable objects of the locally finitely presentable category $\mathcal{Ind}((\mathsf{Ab}_{fp})^{op})$.

To put it another way, $\mathcal{Ind}((\mathsf{Ab}_{fp})^{op})$ is essentially algebraic and in a formal sense has a presentation by certain restricted "partial" operations and equations.

My question is whether a concrete description of $\mathcal{Ind}((\mathsf{Ab}_{fp})^{op})$ is known.

Note that its dual is not the category of profinite abelian groups, since I am considering the finitely presentable rather than the finite abelian groups.

There are many examples where the analogous construction leads to a well-known category e.g. starting with $\mathsf{Set}$ we obtain $\mathcal{Ind}(\mathsf{Set}_{fp}^{op}) \cong \mathsf{BA}$, starting with $\mathsf{DL}$ we obtain $\mathsf{Poset}$, starting with $\mathsf{Vect}(\mathbb{F})$ we obtain the same again, and likewise for join-semilattices with bottom.

Many thanks.

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You wrote: "However one nice property of finitely presentable abelian groups is that they are closed under finite limits, in fact under finite products and SUBALGEBRAS". THen I ask: Is a subgroup $H$ of a finitely presentable abelian group $G$, also finitely presentable? It seem (to me) a no trivial question (as the analogous for the quotient). Where can I find a proof? –  Buschi Sergio Sep 5 '13 at 9:29
    
Edit: I resolved. –  Buschi Sergio Sep 5 '13 at 10:04
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1 Answer

up vote 6 down vote accepted

Very interesting problem! I'd not seen it before, but from what I can make out, it looks as though this category can be described concretely as having for its objects triples $(A, T, i: A \otimes \mathbb{Q}/\mathbb{Z} \to T)$ where $A$ is an abelian group, $T$ is a torsion abelian group, and $i$ is an injective homomorphism. Morphisms are given by pairs of homomorphisms $A \to A'$, $T \to T'$ that are compatible with the injective homomorphisms. (Hence, something like a gluing construction.)

The $Ind$-completion of the opposite of finitely presented abelian groups is equivalent to the category of left exact functors $\Phi: \text{Ab}_{fp} \to \text{Set}$, so I'll start there.

Finitely presentable abelian groups are the same as finitely generated $\mathbb{Z}$-modules. Every such module $M$ is a product of a finite power $\mathbb{Z}^n$ and a finite abelian group $F$, so that by left exactness $\Phi(M)$ can be written in the form $\Phi(\mathbb{Z})^n \times \Phi(F)$. It makes sense to consider the restriction of $\Phi$ to the full subcategories $\text{Ab}_{\text{fin}}$ (of finite abelian groups) and $\text{Pow}(\mathbb{Z})$ (of finite powers of $\mathbb{Z}$). Each of these subcategories is finitely complete, and the full inclusions are left exact, so that $\Phi$ restricts to left exact functors on each of these two subcategories.

Interestingly, both of these subcategories happen to be self-dual. Thus in the first case we may as well consider the category of left exact functors $\text{Ab}_{\text{fin}}^{op} \to Set$, that is to say the $Ind$-completion of $\text{Ab}_{fin}$, which is the category of torsion abelian groups. Hence the first restricted functor is given by $\hom(-, T)$, where $T$ is a torsion abelian group.

In the second case, we may as well consider the category of left exact functors $\text{Pow}(\mathbb{Z})^{op} \to Set$. The domain is the Lawvere theory of abelian groups; left exact functors are product-preserving functors, so such a functor must be of the form $\hom(-, A)$ for some abelian group $A$. In fact, in this case left exact functors coincide with product-preserving functors, essentially because exact sequences in $\text{Pow}(\mathbb{Z})$ split.

The data $(T, A)$ we have thus extracted do not completely characterize $\Phi$ because we have not taken into account how $\text{Pow}(\mathbb{Z})$ and $\text{Ab}_{\text{fin}}$ interact in $\text{Ab}_{fp}$. We need to consider that $\Phi$ preserves the kernel pair of a morphism $f: \mathbb{Z}^n \to F$ mapping to a finite abelian group. It is easy to convince oneself that $\Phi(f): A^n \to \hom(F, T)$ is a group homomorphism. Since $F$ can be decomposed further as a product of cyclic groups, ultimately we find that the left exactness requirement boils down to the fact that $\Phi$ should preserve exact sequences of the form

$$0 \to \mathbb{Z} \stackrel{- \cdot n}{\to} \mathbb{Z} \to \mathbb{Z}_{n}$$

so that we require exactness of

$$0 \to A \stackrel{- \cdot n}{\to} A \to T_{n}$$

where $T_n$ is the subgroup of elements of $T$ annihilated by $n$. This can be equivalently rephrased as saying that the induced map $A \otimes \mathbb{Z}/(n) \to T \otimes \mathbb{Z}/(n)$ is injective (naturally over all $\mathbb{Z}_n$), and the neatest way of encapsulating all is by passing to the (filtered) colimit over all $\mathbb{Z}/(n)$, which leads to the statement that we have an injective map

$$A \otimes \mathbb{Q}/\mathbb{Z} \to T \otimes \mathbb{Q}/\mathbb{Z} \cong T.$$

There are quite a few details to be checked in this analysis, and I don't claim I've checked every one, but I should imagine this is in the right direction.

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Thanks, this is very helpful. –  Rob Myers Sep 4 '13 at 12:00
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Interesting description. Is this the category of (special) sheaves on some nice space (site)? Then the description should come from a decomposition of the space into an open and a closed set. –  Martin Brandenburg Sep 4 '13 at 23:46
    
Would you mind providing some extra detail? I am finding it hard to understand how the finitely presentable abelian groups correspond on the dual side. Also, I'd be interested in any connection with Pontryagin duality. Any help much appreciated! –  Rob Myers Jan 23 at 21:52
    
@RobMyers Sorry, I'm having trouble understanding what your question is exactly. Possibly this is relevant: math.stackexchange.com/questions/126584/… ? Feel free to write me at my email address. –  Todd Trimble Jan 23 at 23:16
    
@Todd Trimble: Thanks, which email address do you mean -- the topological musings one? –  Rob Myers Jan 24 at 15:14
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