MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One way to define the free abelian group on a set $S$ is as $F(S) := \mbox{Hom}_{\text{Set}}(S, \mathbb{Z})$ with the group operation coming by composition with the map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. The problem with this definition is that it's contravariant, whereas it seems to me that $F$ should really be a covariant functor since it should take functions $S \to T$ to homomorphisms $F(S) \to F(T)$. (Or should it?)

One way to get a covariant functor is to actually define $F(S) := \mbox{Hom}_{\mbox{Ab}}( \mbox{Hom}_{\text{Set}}(S, \mathbb{Z}), \mathbb{Z})$. It's interesting to think of the free abelian group as a "double dual" construction, but now I'm not sure that one gets the group operation in a natural way. Which of these definitions is "correct" (if either), and why?

Edit: Let me rephrase my question, then. One way to explicitly describe the free abelian group on a set $S$ is as the direct sum of $|S|$ copies of $\mathbb{Z}$. Is there a categorical description of this construction, i.e. one which doesn't refer to elements, only morphisms?

(If it helps, I'm trying to digest more thoroughly the relationship between chain and cochain groups.)

share|cite|improve this question
The parser seems to be reading the double underscore the wrong way... – Qiaochu Yuan Nov 22 '09 at 23:12
Technically there's a problem with your definition as you need the homomorphisms to be finitely supported -- otherwise S isn't a basis. – Ryan Budney Nov 22 '09 at 23:21
Contravariance you can only get if your maps of sets have finite fibres. Covariance basically comes for free, right? – Ryan Budney Nov 22 '09 at 23:30
Ryan, good point. That means both definitions are wrong, but I would really appreciate an answer of the form "the free abelian group can be obtained as a colimit over such-and-such diagram." – Qiaochu Yuan Nov 22 '09 at 23:35
This isn't phrased the way you want but I think it answers your question in some regard. As abelian groups, chains and cochains are isomorphic provided your set S (above) is finite, by the universal properties of direct sums and products. In general there's a map from chains to cochains but it's not always an isomorphism. Moreover, it's never natural, because the morphisms for chains are covariant in S, and for cochains they're contravariant in S. So in the finite-dimensional setting the thing that differentiates the two is their maps, not the groups. – Ryan Budney Nov 23 '09 at 0:09
up vote 10 down vote accepted

As I understand things, the desire to find a more categorical construction of the free abelian group on a set S is not itself fully in the categorical spirit. Rather, for an object which is defined by a universal mapping property, you only need to convince yourself that it exists; you don't need to be bothered by looking at any particular construction. The point is that anything that you want to know about this object will follow most transparently from the universal mapping property.

In this case, if we have a universal map S -> FreeAb(S) and a map of sets S -> T, we can define a map FreeAb(S) -> FreeAb(T) without "looking under the hood" at how FreeAb's are constructed. Namely, by composing S -> T and T -> FreeAb(T) we get a set map S -> FreeAb(T). By the universal mapping property, this factors through a homomorphism FreeAb(S) -> FreeAb(T), which is what we wanted. Done!

share|cite|improve this answer
we can also note that the forgetful functor Ab->Set admits an adjoint, and we can prove fairly easily that the adjunction is monadic, i.e. Ab is algebraic over Sets, which should give us our free object functor. – Harry Gindi Nov 23 '09 at 4:22
I have absolutely no argument with this answer (which is excellent and to the point). But I think that saying "anything you want to know about this object will follow most transparently from the universal mapping property", while perhaps true of free abelian groups, does not generalize well. For example, free (nonabelian) groups are universal objects in the category of nonabelian groups, and from this universality many things follow, but I would hardly say that "anything" you want to know about free groups is most easily seen in this categorical way. – Danny Calegari Nov 23 '09 at 17:37
It works well in some other examples (e.g. tensor products), and not so well in others (free groups, agreed). There was a role-playing aspect to my response: the OP essentially asked for the "categorically correct" way of thinking about something, so I answered with what I believe to be the correct statement of the categorical philosophy. I do subscribe to this philosophy, but I have subscriptions to other philosophies as well: I certainly do not agree that the best way to look at things is always the most categorical way. Insert Hamlet quote about heaven, earth and Horatio here. – Pete L. Clark Nov 23 '09 at 20:40

To address your parenthetical comment, one can define the homology of a space with coefficients in any object M of any abelian category C by defining the "chain objects" by means of the functor which sends a set S to the coproduct of copies of M indexed on S. When C = Ab, this is usual homology with coefficients; when C = Abop, it's cohomology. This homology is again an object of C, which explains (if you work it out) why cohomology is still covariant in the module of coefficients when viewed as a functor H*(X, –) : Ab → Ab.

share|cite|improve this answer

You can do the construction 'categorically' as follows: using the Representability theorem (given as Thm. 3, Sect. 6, Chap. V in MacLane's book on categories) one can show that the forgetful functor $\mathbf{Grp}\to\mathbf{Sets}$ has a left adjoint $F:\mathbf{Sets}\to\mathbf{Grp}$, which constructs free groups, «without entering in the usual explicit construction», as MacLane himself observes in an example after the theorem. Likewise, you can use that theorem to show that the forget functor $\mathbf{Ab}\to\mathbf{Grp}$ has a left adjoint, which is, of course, the abelianization $\mathrm{ab}:\mathbf{Grp}\to\mathbf{Ab}$.

You are after the composition $\mathrm{ab}\circ f$.

share|cite|improve this answer

Certainly $\\text{Hom}\_\{\\text{Ab}\}(\text{Hom}\_{\text{Set}}(S,\mathbb Z),\mathbb Z)$ is an abelian group, since $\text{Hom}\_{\text{Set}}(S,\mathbb Z)$ is an $\text{Ab}$ is $\text{Ab}$-enriched. But it is not the free abelian group generated by $S$.

One can explicitly construct a free abelian group: it should be the free module generated by $S$, i.e. it should consist of finite linear combinations of elements of $S$. Because the free group on $S$ is defined by $\text{Hom}\_{\text{Ab}}(\text{Free}(S),-) = \text{Hom}_{\text{Set}}(S,\text{Forget}(-))$ (a natural isomorphism; by Yoneda this definition picks out $\text{Free}(S)$ up to isomorphism), and you can chase elements to see that the group of finite linear combinations of elements of $S$ satisfies the definition.

So for example, the free abelian group generated by a countable set $N$ is the countable coproduct of copies of $\mathbb Z$. On the other hand, $\text{Hom}\_{\text{Set}}(N,\mathbb Z)$ is the countable product of copies of $\mathbb Z$, which is an uncountable set and is not free as an abelian group, and $\\text{Hom}\_\{\\text{Ab}\}(\text{Hom}\_{\text{Set}}(N,\mathbb Z),\mathbb Z)$ is also uncountable, I think.

So $\\text{Hom}\_\{\\text{Ab}\}(\text{Hom}\_{\text{Set}}(S,\mathbb Z),\mathbb Z)$ is not equal to $\text{Free}(S)$. Rather, by definition of $\text{Free}$, it is the double dual: $\text{Free}(S)$ embeds canonically into $\\text{Hom}\_\{\\text{Ab}\}(\text{Hom}\_{\text{Set}}(S,\mathbb Z),\mathbb Z)$, but the embedding is not surjective if $S$ is infinite.

share|cite|improve this answer
Thanks. I've rephrased the question. – Qiaochu Yuan Nov 22 '09 at 23:51

Your edited question is:

One way to explicitly describe the free abelian group on a set S is as the direct sum of S copies of . Is there a categorical description of this construction, i.e. one which doesn't refer to elements, only morphisms?

That description is already categorical. The direct sum of abelian groups is the coproduct, so once you are convinced that the free group on a single element exists (i.e. $\mathbb{Z}$) then you have them all. This is implicit in Theo's answer, but I guess he was replying to your original question. More generally, in a variety of algebras (in the sense of universal algebra), once you have the free thing on one generator then you have all of them. The universal property says that we are looking for something such that

$$ \operatorname{Hom}_{\operatorname{Set}}(S,|V|) \cong \operatorname{Hom}_{\operatorname{Alg}}(F(S),V) $$

(where $\operatorname{Alg}$ stands for our variety of algebras) but since $S = \coprod_S \lbrace \star \rbrace$, we already have

$$ \begin{aligned} \operatorname{Hom}_{\operatorname{Set}}(S,|V|) &\cong \operatorname{Hom}_{\operatorname{Set}}(\coprod_S \lbrace \star \rbrace,|V|) \\ &\cong \operatorname{Hom}_{\operatorname{Set}}(\lbrace \star \rbrace,|V|)^S \\ &\cong \operatorname{Hom}_{\operatorname{Alg}}(F(\lbrace \star \rbrace),|V|)^S \\ &\cong \operatorname{Hom}_{\operatorname{Alg}}(\coprod_S F(\lbrace \star \rbrace),|V|) \end{aligned} $$

Hence $F(S) \cong \coprod_S F(\lbrace \star \rbrace)$.

With regard to the issues in the original setting about taking $\operatorname{Hom}_{\operatorname{Set}}(S,|\mathbb{Z}|)$ and then taking homs into that, there is a context in which that works: one has to work with topological abelian groups. Then the coproduct has the inductive topology and the product has the projective topology, and they are dual for any set regardless of size. This is, in fact, used in cohomology theories to make the duality of cohomology and homology work for non-finite CW complexes (and beyond). Or, to be more accurate, to make it even possible that cohomology is dual to homology - there are still more issues which mean that it often isn't but that is for interesting reasons rather than the dull one of failing to take topology into account. (Note that there's nothing special about abelian groups here, this works for any variety of algebras).

For more, particularly the applications to cohomology theories, see the papers by Boardman and by Boardman, Johnson, and Wilson in the Handbook of Algebraic Topology, and my paper with Sarah Whitehouse: The Hunting of the Hopf Ring.

share|cite|improve this answer

There are still strictly speaking elements floating around in the following since we are using indexing sets but maybe it is better? Consider for a set $S$ and an abelian group $A$ the isomorphisms $$Hom_{\mathrm{Set}}(S,UA) \cong Hom_{\mathrm{Set}}(\coprod_S \{*\},UA) \cong \prod_S UA$$ where $U$ is the forgetful functor. Since $U$ creates limits and we want $F$ to be left adjoint to $U$ the free abelian group $F(S)$ is the corepresenting object for the functor sending $A$ to $\prod_S A$. This is really just a reinterpretation of the fact that we want to hit things with the free abelian group functor but maybe it is closer to what you are looking for?

Also I am not completely sure what you mean by your last comment - if you are trying to understand the relationship between homology and cohomology via the chain/cochain complexes you might be after the following?

Theorem Let $R$ be a ring and let $P$ be a chain complex of projective $R$-modules such that each $d(P_i)$ is also projective. Then for every $i$ and every $R$-module $N$, there is a non-canonically split short exact sequence $$ 0 \to Ext^1_R(H_{i-1}(P),N) \to H^i(Hom_R(P,N)) \to Hom_R(H_i(P),N) \to 0$$

So in particular for complexes of free abelian groups one can apply this and it tells you how taking duals alters the cohomology.

I'd like to second Ryan Budney's answer about the difference being the maps in the finite rank setting. Although unless you have a natural setting in which you are considering chains and cochains (for instance singular homology/cohomology) I still don't quite understand exactly what you are after since you can always reindex to move between them.

share|cite|improve this answer
Yes, I want an answer like "take the direct sum of |S| copies of Z," except in more categorical language. As for the other comment, I mean something much more elementary: I am trying to convince myself that chains are not the same thing as cochains. – Qiaochu Yuan Nov 22 '09 at 23:31

You might not find this satisfying, but I think the standard description $\displaystyle{\bigoplus_{s \in S} \mathbb{Z} e_s}$ not only makes the adjunction clear on the level of objects, but also makes it clear that morphisms are respected. If we want to describe maps out of a free abelian group, it is reasonable to specify them on the explicit generators that are given to us.

You could "define" it as the reduced homology of a wedge of circles indexed by S, but I don't know if that buys you anything. I suppose if you replace "homology" with "fundamental group" you get a better-looking definition of free group than the one with reduced words, so that's one way of saying it's more canonical. Again, the description of a free group as a coproduct of copies of Z with specific generators doesn't seem to be so problematic from a categorical standpoint, since maps from Z are in bijection with elements of a group.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.