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Must the natural quotient group of the inverse limit of a sequence of nested discrete free monoids have topological dimension zero?

The question might well be open, but I would be grateful for news to the contrary, history of the problem, or other feedback.

To make the question precise, suppose $M_{1} \subset M_{2}...$ is a sequence of discrete free monoids such that $M_{n}$ is generated by a set $L_n \cup L_n^{-1}$. Suppose we have retraction epimorphisms $r_{n}:M_{n+1} \rightarrow M_{n}$ (fixing $M_{n}$ pointwise). Suppose $M=lim_{\leftarrow}M_{n}$ is the inverse limit of the spaces $M_{n}$.

Let $G_{n}$ denote the discrete canonical free group determined by $M_{n}$. Let $q:M \rightarrow lim_{\leftarrow}G_{n}$ denote the canonical homomorphism, and let $G=q(M) \subset lim_{\leftarrow}G_{n}$.

If G has the quotient topology must G have topological dimension zero?

To see why the question hard, by construction, $G$ naturally continuously injects into $lim_{\leftarrow}G_{n}$, and thus $G$ is Hausdorff and totally disconnected. However the injection $G \rightarrow lim_{\leftarrow}G_{n}$ is unlikely to be a topological embedding.

(For the simplest counterexample, if $G_n$ is the free group on $x_{1},x_{2},...,{x_n}$ then G is the fundamental group of the Hawaiian earring. Moreover examples such as Erdos space reveal that a totally disconnected Hausdorff space can have nonzero dimension.)

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