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This question is quite specific, but it may admit answers in more general contexts.

Consider a subset $\Lambda \subset D^2$ where $D^2$ is the two dimensional disk.

We consider in $\Lambda$ an equivalence relation such that the equivalence class of each point is a contractible compact set.

Assume that the quotient map $p: \Lambda \to \Lambda / \sim$ which is continuous has as image a cantor set.

The question is: If we extend the equivalence relation to the whole disk by considering for each point $x\in D^2 \backslash \Lambda$ the equivalence class is the singleton $\{x\}$ do we have that the proyection to the quotient of the whole disk is homeomorphic to the disk?

If the answer is negative, can we ask more to the equivalence classes in $\Lambda$ in order to have the result?

Maybe the question is trivial or well known, but I could not find either a reference nor an answer by myself.

EDIT: In view of Franklin's answer. I am supposing that $\Lambda$ is contained in the interior of the disk (which I am assuming closed).

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Do you know the Kline sphere characterization? I think there is another characterization only mentioning points which should easily apply. Maybe that removing a point changes the local fundamental group? – Ben Wieland Jul 14 '10 at 0:59
I don´t understand your claim. Here I am not removing points, just collapsing some sets. For example, imagine they are arcs, each one of them, doesn´t change the topology, but here we are collapsing non-countably many and so I cannot see how it works. – rpotrie Jul 14 '10 at 10:15
If the quotient satisfied the hypotheses of the characterization, the characterization would conclude that the quotient is a disk. – Ben Wieland Jul 14 '10 at 17:41
up vote 3 down vote accepted

I think you may find the Bing shrinking criterion useful.

First, assume $\Lambda$ itself be closed (hence compact) in $D$. More generally, equivalence classes can form a so-called upper semi-continuous decomposition of your compact initial space $X$, namely one such that $X/\sim$ is Hausdorff (necessary anyway), making the quotient map $p$ closed.

Bing shrinking criterion : $p:X\to X/\sim$ is a uniform limit of homeomorphisms iff for any $\epsilon>0$, there is an homeomorphism $h_\epsilon$ of $X$ that send any equivalence class into a set of diameter $<\epsilon$ which is moreover contained in the $\epsilon$-neighborhood of the original class.

And one proof (by Robert Edwards) is a beautifully simple application of Baire's theorem, which works for any usc decomposition of a compact metric space. See this 1979 Bourbaki talk (page 10) by Edwards himself.

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Thanks for the answer and the reference! Just one question to finish the proof: it seems as if one could weaken the hypothesis of equivalence classes being contractible, does it work if we assume that they are continua? – rpotrie Jul 14 '10 at 12:58
Not all continua will do. They must be cellular, i.e. decreasing intersections of cells (sets homeomorphic to a closed disk). Even in dimension 2, there are strange beasts like the pseudo-arc among cellular continua. See and the reviews citing this. For modern accounts of continuum theory and decomposition spaces of manifolds, you can try the books by Nadler and Daverman with these titles, although Bing's collected works are also great. – BS. Jul 14 '10 at 13:56
Thanks a lot! Decreasing intersection of cells work perfect for me! Thanks again for the references. – rpotrie Jul 14 '10 at 15:50
Let me add that equivalence classes must be cellular if the quotient space is to be a manifold, which was the context of your question. – BS. Jul 14 '10 at 17:29

Imagine that the Cantor set is on one diameter and that $\Lambda$ consists of the vertical cords passing through the Cantor. After collapsing you get a space that have some points that removing them makes it disconnected. Therefore is not homeomorphic to the disc.

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Thanks. Sorry, I must edit the question, I am thinking of $\Lambda$ contained in the interior of the disk. – rpotrie Jul 13 '10 at 19:15

The equivalence classes should be closed subsets of the disk to make the quotient Hausdorff.

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You are right. I will correct that. – rpotrie Jul 13 '10 at 23:05

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