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Let $X$ be a locally compact metric ANR (or, if preferred, a locally compact simplicial complex). If needed, assume that $X$ has finitely many ends or is of finite dimension. My question is:

What are the necessary conditions for the Freudenthal compactification of $X$ to be a metric ANR?

In the literature I have found some sufficient conditions (though up to date I was not able to dig through the papers I refer to and compare the conditions), like:

  1. forward tameness (Hughes, B., Ranicki, A.: Ends of Complexes, Chapter 7.)
  2. docility at infinity (Sher, R. B.: Docility at infinity and compactifications of ANR's)
  3. $\mathcal{C}_p$-movability at infinity (Cerin, Z.: $\mathcal{C}_p$-movable at infinity spaces, compact ANR divisors and property $UVW^n$)
  4. $SUV^\infty$ (Sher, R. B.: A theory of absolute proper retracts)

A full characterization of $X$'s with the property I ask for is probably difficult. An analogous question for the one-point compactification is stated open (Problem 79SC6) in the book J. van Mill, G.M. Reed: Open Problems in Topology. At least in the case of finitely many ends these two problems look at least close to being equivalent. The book has some references to the papers of Dydak ('On $LC^n$ divisors' and 'On maps preserving $LC^n$ divisors'), which probably give an answer in the finite dimensional case, but which at the moment I still do not understand (if some short exposition is possible, it would be welcome).

Perhaps for me even better than the answer to the first question would be a good answer to the following one:

What are some simple examples of $X$ whose Freudenthal compactification is not an ANR and why is it so?

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I think gt.geometric-topology and at.algebraic-topology would be relevant tags. –  Sergey Melikhov Apr 19 '12 at 16:53

1 Answer 1

up vote 2 down vote accepted

Regarding simple examples: take $X$ to be the infinite mapping telescope of an inverse sequence of connected compact polyhedra $P_i$ (so that $X$ has one end) such that for some $j$, the inverse sequence of the groups $G_i=H_j(P_i)$ does not satisfy the Mittag-Leffler condition or its inverse limit does not inject in any $G_i$. Then the Freudenthal compactification, which in this case is the same as the one-point compactification $X^+$, is not an ANR.

The Mittag-Leffler condition is not satisfied for instance if each $P_i=S^j$ and each bonding map $P_{i+1}\to P_i$ is a degree $2$ map ($j>0$). The inverse limit does not inject in any $G_i$ for instance if each $P_i$ is the $j$-fold suspension $S^{j-1}*[2^i]$ over the discrete space of cardinality $2^i$ (here $j>0$ in order to have $P_i$ connected), and each bonding map $P_{i+1}\to P_i$ is the suspension over a trivial $2$-fold cover.

More generally one can take the $P_i$ to be disconnected, but such that the inverse limit of $\pi_0(P_i)$ injects into some $\pi_0(P_i)$. Then $X$ has finitely many ends (though there might be infinitely many proper homotopy classes of proper maps $[0,\infty)\to X$ if $\pi_1(P_i)$ do not satisify the Mittag-Leffler condition). So if the Freudenthal compactification of $X$ is an ANR, then $X^+$ is an ANR.

So why is $X^+$ not an ANR if the $G_i$ do not satisfy the Mittag-Leffler condition or their inverse limit does not inject in any $G_i$? This follows from Dydak's necessary and sufficient condition for $X^+$ to be ANR, where $X$ is the infinite mapping telescope of an inverse sequence of compact polyhedra $P_i$, in terms of the homology groups and the fundamental groups of the $P_i$. I've tried to understand this result, and have written up a somewhat different proof of the homological part (which perhaps also counts as a short exposition that you request), see Theorem 6.12 in http://arxiv.org/abs/0812.1407. Theorem 3.12 also gives a similar necessary and sufficient condition for the mapping telescope $X$ to be forward tame, in terms of the homotopy groups of the $P_i$. See also Lemma 3.4 for a quick explanation of some terms.

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