Two responses to a recent question.........DMD
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Subject: Re: response to response
From: Tom Goodwillie <tomg@math.brown.edu>
Date: Tue, 10 May 2005 10:20:54 -0400

It's not true for n=1, if the circle is a nice compact space and the 
universal covering map from the open 1-disk to the circle is a nice map.

Tom Goodwillie

>>>   >> From: Andre Henriques <andrhenr@math.mit.edu>
>>>   >> Date: Thu, 5 May 2005 23:11:28 -0400 (EDT)
>>>   >>
>>>   >> Here's a harder question
>>>   >> related to that of Johannes Huebschmann
>>>   >> whose answer I'd be interested to know:
>>>   >>
>>>   >> Let D be the open n-disk and \bar D be the closed n-disk.
>>>   >> Let f be a 'nice' map from D to some 'nice' compact space X.
>>>   >> Is there a compactification B of D such that f extends to B and
>>>   >> such that the pair (B,D) is homeomorphic to the pair (\bar D,D)?
>>>   >>
>>>   >> Feel free to substitute 'nice' by any notion that you find convenient.
>>>   >> Andre Henriques
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Subject: Re: response to response
From: Yuri Turygin <turygin@math.ufl.edu>
Date: Wed, 11 May 2005 05:43:09 -0400 (EDT)

The answer to Andre's question:

A continuous map f:D\to X extends to \bar f:\bar D\to X if and only if
f:D\to X is uniformly continuous.

Pf: If f extends to \bar f, then DONE. Now, suppope f:D\to X is uniformly
continuous. For each point on the boundary x\in \bar D\setminus D define
\bar f(x):=lim(f(x_n)), where the sequence {x_n} is such that lim x_n = x.
Now check that bar f is well defined. Suppose there exist {x_n} and {x'_n}
such that lim x_n=lim x'_n = x, but lim(f(x_n))\neq lim(f(x'_n)). Then for
any \delta you will find \delta-close to each other x_n and x'_n whose
images f(x_n) and f(x'_n) will be \epsilon-far, where \epsilon is, say,
d_X (lim(x_n), lim(x'_n))/2 (here d_X is a metric on X, but X doesn't
have to be a metrizable compactum for my claim to be still true. My proof
will also work with non-metric version of uniform continuity.
X has only to be a Hausdorff compactum).

Best,
Yuri Turygin