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I know the limits in the categories of groups, abelian groups and topological spaces and was wondering about the same thing.

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2 Answers 2

up vote 14 down vote accepted

The product of topological groups is simply the product of the underlying groups with the product topology. The universal property is easily verified. The coproduct topology is more complicated. Again, the underlying group is just the coproduct of the underlying groups, which is also called the free product. Here is a description of the topology:

Let $(G_i) (i \in I)$ our family of topological groups. Then consider the class of all topological groups $H$, such that there is a surjective group homomorphism $G:=\coprod_{i \in I} G_i \to H$, such that the composites $G_i \to H$ are continuous. Notice that this class is essentially a set. Namely, consider the set of all quotient groups $G / N$ endowed with some topology. In particular, we may define the topology on $G$ to be the initial one with respect to all maps $G \to H$ as above. Then $G$ is a topological group. For example, the inversion map $G \to G$ is continuous since the composite with every $G \to H$ as above is the composite with $G \to H$ and the inversion map $H \to H$. The universal property is also easy to verify.

This works in general: If $\tau$ is some type of algebraic structures, then the category of topological $\tau$-algebras is complete and cocomplete, and the forgetful functor to $\tau$-algebras preserves and creates limits and colimits. You can also prove this by applying Freyd's representabilty theorem. In the above example, I have written down a solution set.

However, it is hard to describe this topology explicitely (and thus, for instance, proving separation axioms). See for example this or that article.

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thanks, good answer! –  mayonnaise Sep 21 '10 at 15:28
    
Dear Martin, I don't understand your use of Freyd's representability theorem in this context, nor the role of the solution set in this theorem. Would you be so kind to explain them? Also: do you know a reference for the fact that the category of topological groups is cocomplete? –  a.r. Oct 1 '10 at 9:23
    
MartinBrandenburg, @AgustíRoig - the other answer posted here intends to ping you two. –  anon Sep 1 '12 at 21:27

@Martin: @Agusti Just come across your answer, Martin, which rightly pointed out the use of Freyd representability to get a general result. In fact this was used in essence in the paper

R. Brown and J.P.L. Hardy, ``Topological groupoids I: universal constructions'', Math. Nachr. 71 (1976) 273-286.

to give colimits of topological groupoids, and so in particular topological groups.

As you say, it is difficult to get hold of specific properties of this topology; the construction is defined by its universal property, and that is all.

However, in the case of free topological groups, there are some quite specific exploration of the properties, with papers by Markov, Graev, and S.A. Morris, P. Nickolas.

Here are two papers which should be relevant to the discussion:

Nickolas, Peter, Coproducts of abelian topological groups. Topology Appl. 120 (2002), no. 3, 403–426.

Nickolas, Peter, A Kurosh subgroup theorem for topological groups. Proc. London Math. Soc. (3) 42 (1981), no. 3, 461–477.

Another way of dealing with this problem is to work in a convenient category of spaces, usually the category of compactly generated spaces, in which the product of identification maps is an identification map.

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You may be interested to know @name does not ping other users when placed in answers. I have put a comment on the other answer instead - Martin will get the ping without an @ because the comment is on his answer, and I put an explicit @ for Agusti. (The system only allows one @-caused ping per comment.) –  anon Sep 1 '12 at 21:30
    
Thank you, Ronnie. It was some time ago, but never mind. :-) –  a.r. Sep 1 '12 at 21:36

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