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Let $X$ be a compact metric, second countable space with finite covering dimension. Let $A,B$ be two closed subsets of $X$. Assume that $h:A\to B$ is a homeomorphism. Is it possible to extend $h$ to a homeomorphism $\tilde{h}: X\to X$?

If not, are there at least some conditions under which it is possible?

Thanks!

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    $\begingroup$ Let $X=[-1,1]$, let $A=\{0\}$ and let $B=\{1\}$. There is no homeomorphisms of $X$ that maps $0$ into $1$. $\endgroup$ – erz Oct 6 '18 at 5:57
  • $\begingroup$ If $X$ is homogenous, you can do it for two singletons. $\endgroup$ – Henno Brandsma Oct 6 '18 at 8:57
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    $\begingroup$ Knot theory deals with waysto embed $S^1$ into $\mathbb{R}^3$. Such embeddings can be distinguished by their complement. So there are plenty of knots where their complements are not homeomorphic. Thus there cannot be a homeomorphism as in this question. $\endgroup$ – HenrikRüping Oct 6 '18 at 13:55
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In the "positive" direction, there exist some deep results (called Z-set unknotting theorems) on extensions of homeomorphisms between Z-sets of Menger manifolds, see Theorem 4.1.18 in the book "Inverse Spectra" of Chigogidze.

In the simplest form the Z-set Unknotting Theorem (proved by Bestvina) says that a homeomorphism $h:A\to B$ between two $Z$-sets of the $n$-dimensional Menger cube $M$ extends to a homeomorphism of $M$.

A closed subset $A$ of a compact topological space $X$ is called a $Z$-set in $M$ if the set of maps $M\to M\setminus A$ is dense in the space of all self-maps of $M$, endowed with the compact-open topology. Each Z-set is nowhere dense in $M$.

If $M$ is zero-dimensional, then a closed subset $A\subset M$ is a $Z$-set if and only if it is nowhere dense. In particular, any homeomorphism between closed nowhere dense subsets of the Cantor set $M$ extends to a homeomorphism of $M$.

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As you can see from the comments the answer is: hardly ever.

As mentioned above the case of one-point sets necessitates the space being homogeneous. But that is not enough, say in $\mathbb{R}$ when you map $\{1,2,3\}$ to itself by interchanging $1$ and $2$.

You may also want to have a look at Antoine's necklace.

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