Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Call a "blot" set, which is the closure of its interior, the boundary is locally connected, and when you remove boundary blot remains connected. Suppose that there is a blot on the surface of the n-dimensional sphere. Is it true that every homeomorphism of the blots on itself extends to a homeomorphism of the sphere on itself? If this is not true, can it be imposed on the blot any additional reasonable conditions (eg, smoothness) to the statement was true?

share|improve this question
    
No. For example, let $X \subset S^3$ be a closed tubular neighborhood of the unknot. Thus $X$ is a solid torus (ie $D^2 \times S^1$) and $S^3 \setminus X$ is the interior of a solid torus. Let $f : X \rightarrow X$ be the homeomorphism obtained by cutting $X$ open along the disc $D^2 \times 1$, giving the cut open space a full twist, and regluing. Then it is easy to see that $f$ does not extend to a homeomorphism of $S^3$. –  Andy Putman Dec 15 '10 at 0:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.