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For any set $S$ one can consider the free abelian group $\mathbb{Z}[S]$ generated by this set. Now suppose, there is a topology on $S$ given. Is it possible to find a topology on $\mathbb{Z}[S]$ in such a way, that:

i) The map $S\rightarrow \mathbb{Z}[S]$ is a homeomorphism onto the image.

ii) The addition and the inverse map are continuous

And if it is possible, is this topology unique?

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My recollection is that such a topology on $\mathbb{Z}[S]$ does exist (and hence is unique by Fabrizio Polo's argument below), but that it is rather subtle and difficult to construct. I seem to remember that it depends on what exact category of topological spaces you are working in, e.g. compactly generated weakly Hausdorff spaces? vanilla topological spaces? etc. I think that it is easier to show that there exists some "free abelian group" generated by a topological space in the sense that it has the universal property. It's harder to show that it's underlying set is $\mathbb{Z}[S]$. –  Chris Schommer-Pries Mar 30 '10 at 13:48
    
I added the algebraic topology tag because I think aspects of this question come up in that context. Some algebraic topologists might know an answer to this. –  Chris Schommer-Pries Mar 30 '10 at 13:55
    
Your comment reminds me of the complete group algebra of a profinite group G. This object is frequently used in algebraic number theory. If $\mathcal{O}$ is a, say, coefficient ring, meaning a complete, Noetherian local ring with finite residue field, and $G$ a profinite group (so, as you say, a very special space), then there exists a compact topological ring $\mathcal{O}[[G]]$ and a continuous map $G\rightarrow\mathcal{O}[[G]]$ that parametrizes continuous homomorphisms from $G$ into the group of units of any coefficient ring. –  Keenan Kidwell Mar 30 '10 at 14:02
    
Chris Schomer-Preis' comment made me think a little more. I was under the impression that my construction worked at least when $S$ was something nice like compact metric. If it goes wrong with a stranger space $S$ (as suggested by Dominguez below) then it would be really interesting to see some examples. –  Fabrizio Polo Mar 30 '10 at 14:06
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Yes, algebraic topologists do use this construction regularly, but when S is "nice" (a CW complex or something). The remarkable Dold-Thom theorem is essentially that the homotopy groups of Z[S] are the homology groups of S! –  Dev Sinha Mar 30 '10 at 21:19
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4 Answers 4

up vote 7 down vote accepted

I don't know if such a topology is unique, but it exists if and only if $S$ is completely regular. This includes locally compact hausdorff spaces and CW complexes.

With Freyd's Adjoint Functor Theorem, it can be shown that the forgetful functor from abelian top. groups to top. spaces has a left adjoint. This is essentially the same proof as in the discrete case. Explicitely, $\mathbb{Z}[S]$, the free abelian top. group over the top. space $S$, is the usual free abelian group endowed with the weak topology for all homomorphisms $\mathbb{Z}[S] \to A$, such that $S \to \mathbb{Z}[S] \to A$ is continuous. Here, $A$ is an arbitrary abelian top. group. In order to show that this topology exists, we may assume that $A$ is, as a group, a quotient of $\mathbb{Z}[S]$, so that these $A$ form a set. But the description of the topology does not change and even without Freyd's Theorem it is easy to see that $\mathbb{Z}[S]$ thus becomes an abelian top. group satisfying the desired universal property.

Now I claim that the three assertions

  1. $S \to \mathbb{Z}[S]$ is a homeomorphism onto its image.

  2. $S$ is a subspace of an abelian top. group.

  3. $S$ is completely regular.

are actually equivalent!

1) implies 2), that's clear. Now assume 2), thus $S \subseteq A$ for some top. abelian group. Extend the inclusion $S \to A$ to a continuous homomorphism $\mathbb{Z}[S] \to A$. Every open subset of $S$ can be extended to an open subset of $A$. Pull it back to $\mathbb{Z}[S]$. This is an open subset of $\mathbb{Z}[S]$ which restricts to the given oben subset of $S$. This proves 1).

2) implies 3), this follows from the fact that every topological group is completely regular and subspaces of completely regular spaces are obviously completely regular.

Finally assume 3), i.e. $S$ carries the initial topology with respect to all continuous functions $S \to \mathbb{R}$. Endow $\mathbb{R}^{(S)}$ with the initial topology with respect to all homomorphisms $\mathbb{R}^{(S)} \to \mathbb{R}$, such that the restriction $S \to \mathbb{R}$ is continuous. Then $\mathbb{R}^{(S)}$ is an abelian topological group and $S \to \mathbb{R}^{(S)}$ is an embedding, thus 2).

I also believe that (but cannot prove)

  • If $S$ is hausdorff and completely regular, $\mathbb{Z}[S]$ is hausdorff.

In another comment, it was suggested to endow $\mathbb{Z}[S]$ with the final topology with respect to $S \to \mathbb{Z}[S]$. But this does not even yield a translation invariant topology: If $S=\{a,b\}$ with the only nontrivial open subset $\{a\}$, then $\{a\}$ is open in $\mathbb{Z}[S]$, but $\{b\}$ is not.

But maybe, if $S$ is a completely regular space, the topology of $\mathbb{Z}[S]$ used above is the final topology?

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Free topologican groups, abelian or not, have been studied since the 40s by Markov and Graev (the original references are somewhat obscure). You need your space $S$ to be completely regular for the construction to work. There is a survey by O. V. Sipacheva (THE TOPOLOGY OF FREE TOPOLOGICAL GROUPS, Journal of Mathematical Sciences, Vol. 131, No. 4, 2005) which is maybe too thorough if you just want to get a feeling of this concept, but the first two or three pages contain interesting and easy-to-grasp information on this class of topological groups. The construction itself is not that difficult, and the topology is indeed unique, but they are sometimes difficult to delve into, especially the non-Abelian ones, and a very valuable source of counterexamples.

[edit] I have read the question more carefully and it seems to amount to the following: For a (well behaved) topological space $S$, can one endow the free group $\mathbb Z[S]$ with a group topology such that $S$ becomes a topological subspace of this group? If so, is this construction unique? The answer to the first question is yes, as I have just said. Moreover, $S$ becomes a closed subspace of $\mathbb Z[S].$ I'm not sure about the uniqueness, though. The universal property which characterizes the free topological abelian group is, as long as I remember, the following: it is the only group topology on the free abelian group for which the inclusion mapping $S\to \mathbb Z[S]$ becomes a topological embedding and such that for every continuous mapping $f:S\to G,$ where $G$ is a topological Hausdorff abelian group, the unique homomorphism which extends $f$ to $\mathbb Z[S]$ becomes continuous. This is at first sight a stronger property than the one contained in your question.

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Pestov's review (MR2056625) of the paper is very nice - ams.org/mathscinet-getitem?mr=2056625 –  François G. Dorais Mar 30 '10 at 17:54
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There is a unique topology on $\mathbb{Z}[S]$ for which the map $S\rightarrow\mathbb{Z}[S]$ is continuous such that $\mathbb{Z}[S]$ has the above-mentioned universal property. It is the final topology with respect to the injective map $S\rightarrow\mathbb{Z}[S]$. Any map, continuous or not, $S\rightarrow A$, $A$ an abelian group induces a unique homomorphism of groups $\mathbb{Z}[S]\rightarrow A$ such that $S\rightarrow\mathbb{Z}[S]\rightarrow A=S\rightarrow A$. In particular, if $A$ is an abelian topological group, and $S\rightarrow A$ continuous, then this implies that $\mathbb{Z}[S]\rightarrow A$ is continuous. As I noted in my comment, however, it is not at all clear to me that the group operations on $\mathbb{Z}[S]$ will be continuous with this topology.

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If one considers a discrete set $S$, the topology on $\mathbb{Z}[S]$ shoudl also be discrete. But the map $\mathbb{Z}[S]\rightarrow \mathbb{Z}[S]$ is not continuous, where the source is equipped with the final topology and the target is equipped with the discrete topology. –  HenrikRüping Mar 30 '10 at 14:38
    
In addition to $S \to \mathbb{Z}[S]$, I would take all of its multiples and translates. (You probably need more, but the single map $S \to \mathbb{Z}[S]$ will clearly not do.) –  François G. Dorais Mar 30 '10 at 14:44
    
@François: If one considers $S=\{a\}$ then the resulting final topology w.r.t. all those maps is just the cofinite topology. Hence this also doesn't suffice to make the map to $\mathbb{Z}$ equipped with the discrete topology continuous. –  HenrikRüping Mar 30 '10 at 14:52
    
@Henrik: I'm not understanding what you're saying. If $S$ is discrete then the final topology on $\mathbb{Z}[S]$ is also discrete. –  Keenan Kidwell Mar 30 '10 at 15:36
    
@Henrik: I get the discrete topology. The final topology is the finest topology which makes all the maps continuous. –  François G. Dorais Mar 30 '10 at 15:36
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The answer is yes. This follows from the fact that the group ${\mathbb Z}[S]$ should satisfy a universal property:

Let $A$ be any topological Abelian group and let $f: S \to A$ be a continuous function. Then $f$ can be extended uniquely to a continuous homomorphism ${\mathbb Z}[S] \to A.$

Suppose $G$ and $H$ are two copies of ${\mathbb Z}[S]$ with possibly different topologies. $G$ and $H$ both contain copies of $S.$ Applying the above unviersal property in two directions allows us to see that the identity from $G$ to $H$ is a homeomorphism. More precisely, we get a map $G \to H$ and another $H \to G.$ Composing them one way gives us a map $G \to G$ which must be the identity by the universal property. Composing them the other way gives as a map $H \to H$ which again must be the identity. So the two maps are inverses and each is a homeomorphism.

[Edit: adding the existence portion of the argument]

Consider $(S \cup S^{-1} \cup \{1 \})^{\mathbb Z},$ where $S^{-1}$ is just another ``copy" of $S.$ An element $s \in S$ shall correspond to $s^{-1}$ in $S^{-1}.$ This is a topological space, though I can't remember what topology to put on it. Let $X$ be the subset consisting of all elements with only finitely many coordinates different from $1.$

Now we quotient $X$ in a technically complicated but conceptually simple way. If $x \in X$ has $x_i = s$ and $x_{i+1} = s^{-1}$ for some $s \in S.$ Then we identify $x$ with $y$ where $y=x$ except at places $i,$ and $i+1$ where $y_i = 1 = y_{i+1}.$ We do the same if the $s,s^{-1}$ appear in the other order. Also, if $\sigma$ is a permutation of ${\mathbb Z}$ then we identify $x$ with the $y$ where $y_i = x_{\sigma(i)}.$

To define a group operation, we take two infinite "words", slide them so that they have disjoint "support" and concatenate. (Writing down the details here is harder than letting the reader guess what I mean.)

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This is not a complete argument, yet. You have shown that if there exists a topology on $\mathbb{Z}[S]$ such that with this topology this topological group satisfies the universal property, then it is unique. That does not prove that there exists such a topology in the first place. –  Chris Schommer-Pries Mar 30 '10 at 13:42
    
One way to get the universal property is to endow $\mathbb{Z}[S]$ with the final topology with respect to the map $j:S\rightarrow\mathbb{Z}[S]$. Then any continuous $f:S\rightarrow A$, $A$ an abelian topological group, induces a unique homomorphism $\varphi:\mathbb{Z}[S]\rightarrow A$ such that $\varphi\circ j= f$. This means that $\varphi$ is continuous by definition of the final topology. However, it's not at all clear to me that this will necessarily make $\mathbb{Z}[G]$ into a topological group. I feel like it probably won't in general. –  Keenan Kidwell Mar 30 '10 at 13:52
    
OK so you are suggesting topological abelian groups as test objects and you want to consider the finest topology on $\mathbb{Z}[S]$, that is compatible with +,- and that makes $S\rightarrow \mathbb{Z}[S]$ continuous. OK both of those properties look like they are preserved by refinements ,i.e. given any collection of topologies, their common refinement also satisfies those properties. So the only remaining question is, whether i) is true. –  HenrikRüping Mar 30 '10 at 13:54
    
(i) is true because $j$ is injective, so $j^{-1}(j(U))=U$, whence, if $U$ is open in $S$, $j(U)$ is open in $\mathbb{Z}[S]$. So $j$ is an open embedding. –  Keenan Kidwell Mar 30 '10 at 13:56
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@Keenan: not every injective continuous map is open, so one cannot say that images of open sets are again open. And basically i) can be reformulated as: $j$ is open. I don't see a proof in this comment. –  HenrikRüping Mar 30 '10 at 14:02
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