Subject:
Re: follow-up question
From:
Tom Goodwillie
Date:
Thu, 12 May 2005 22:46:55 -0400
To:
Don Davis
>
>
> Assumption: 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.
>
> Assertion: $X$ is homeomorphic to $D^n$.
>
> As the 2nd example of Kari Ragnarsson's answer (coarsening of topology on the
> boundary) shows, the assertion is not true in general. To exclude such cases,
> add the assumption that $f$ is open. Does the assertion follow then?
If f is open then f must be one to one (so a homeomorphism). In fact, suppose that f(p)=f(q). The points p and q must be in the boundary S^{n-1}. Let U be an open set in D^n containing p but disjoint from a neighborhood of q. Then f^{-1}(f(U)) is open, contains q, but contains no points near q outside S^{n-1} because it contains no points outside S^{n-1} except those in U. Contradiction.
So maybe "open" is not the hypothesis intended.
On the other hand, if we substitute "closed" then here is a counterexample: Let B be a subspace of S3 homeomorphic to D3 such that its complement in S3 is not simply connected -- for example its boundary might be the Alexander horned sphere. Let X be the quotient space D4/B and let f:D4-->X be the quotient map. X is not homeomorphic to D4 because the set S3/B of points at which X is not locally homeomorphic to R4 is not homeomorphic to S3.
Tom Goodwillie