ConstructÂ two homeomorphic closed subsets of some Euclidean space such that their complements are not homeomorphic, but show that if are closed homeomorphic subsets of such that lies in and lies in , their complements in are homeomorphic.

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Take the unknot and the trefoil knot in . Both are obviously homeomorphic, but the knot group of the unknot (that is, the fundamental group of its complement) is isomorphic to , while the knot group of the trefoil is isomorphic to , the braid group on strands.

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It’s even possible to do this in . Take the union of the unit circle and the origin, the union of the unit circle and a point outside of it. The two are obviously homeomorphic but the complement of has a contractible connected component, while the complement of does not.

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To understand the following solution follow this steps (we will draw a simple example): Let . Draw in the horizontal axis . Draw in the vertical axis .

Now, we know that we have a homeomorphism . This map is in particular a bijection, use it to draw the points in of the form , call this set and notice that it also consists of a line and a point. Look at the picture, the claim is that the complement of is homeomorphic to the complement of , which, by symmetry, is homeomorphic to the complement of .

This is pretty credible: See how is the graphic of , since is closed we can use Tietze’s theorem to extend to all and then moving the graphic of this extension up and down we can cover all the plane.

Let us prove this formally, so now we are in the general case. Observe that can be seen as a map . Since is closed, by Tietze’s theorem there is a lift . Define a map from to by

, which is clearly continuous. It is easy to see that restricting and corestricting it, we get a continuous bijection between the complement of and the complement of . Since it is injective, by invariance of domain, it must also be open and thus it is a homeomorphism.

This argument shows that a homeomorphism between and can be lifted to a homeomorphism of the intire space .

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You don’t need to appeal to invariance of domain, since $\bar f$ has an explicit inverse!

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