Subject: Re: three postings
From: "Prof. A. R. Shastri" <ars@math.iitb.ac.in>
Date: Thu, 5 May 2005 19:07:16 +0530 (IST)

 No, you can take $X$ to be the quotient space of the closed 2-ball where 
an arc on the boudary is collapsed to a single point, and f to be the 
quotient map. 

Anant Shastri


On Thu, 5 May 2005, Don Davis wrote:


>> Three postings: A question and two announcements of
>> conference/lecture series..........DMD
>> ____________________________________________________________
>> 
>> Subject: Question
>> From: Johannes Huebschmann <Johannes.Huebschmann@math.univ-lille1.fr>
>> Date: Wed, 4 May 2005 18:38:38 +0200 (CEST)
>> 
>> Let f be a surjective continuous map from a closed n-disk onto a space X
>> whose restriction to the interior of the disk is a homeomorphism onto its
>> image and which has the property that the pre-image of any point of X is
>> a contractible space. Does this imply that f is a homeomorphism?
>> 
>> Regards
>> 
>> Johannes Huebschmann
>> 
>> 

Subject: Re: Question
From: Kari Ragnarsson <kari@maths.abdn.ac.uk>
Date: Thu, 5 May 2005 14:51:13 +0100 (BST)

I believe the answer is no.

As a counterexample, take n=2 and, for convenience, replace the closed
2-disk with the homeomorphic closed unit square I x I. For X, take the
triangle in R2 given in (x,y)-coordinates by 0 <= y <= x <= 1. The map f:
I x I --> X,
given by
  f(x,y) = (x,xy)
satisfies the conditions of the question, but is not a homeomorphism since
the preimage of (0,0) is the unit interval {0} x I on the y-axis.

However, the spaces are still homoemorphic in this case. A funkier example
where the spaces are not homemorphic is given by letting X agree with the
closed n-disk D^n as point sets. We give X the subtopology of D^n
consisting of those open sets in D^n containing all or none of the
boundary. The identity map f: D^n -> X satisfies the conditions in the
question but the spaces are not homeomorphic.

Regards,
Kari

Subject: Re: three postings
From: Jeff Strom <jeff.strom@wmich.edu>
Date: Thu, 05 May 2005 12:55:14 -0400

No.  Let  I  be a (small) closed interval contained in  S1,
and let

  f:  D2 -->  D2/I

be the standard quotient map.  Clearly the point preimages are
either points or  I, which are contractible.  Let  E = f(Int(D2)).
Since  I \cap Int(D2)  is empty,  f  induces a bijection
Int(D2) --> E.  Since  Int(D2)  is open in  D2,  it is not hard to
show that this (continuous) bijection is also an open map
(a set  U  is open in  Int(D2)  if and only if it is open in  D2,
so  f^{-1} f (U) = U is simultaneously open in  Int(D2) and in
D2).

On the other hand,  f  is not injective, so it cannot be a homeomorphism.

Jeff