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Are there two retract subsets $A,B$ of $S^{2}$ with the following property:

$A,B$ are homeomorphic but two pairs $(S^{2},A)$ and $(S^{2},B)$ are not homeomorphic.

The same question for $S^{n}$

Ali Taghavi
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1 Answers1

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I will identify $S^2$ with the Riemann sphere, the 1-point compactification of the complex plane. Start with the graph $G\subset {\mathbb C}$, which is the union of the four segments connecting $0$ with the points $\pm 1$ and $\pm i$. Now let $A$ be the union of $G$ with the two small disks centered at the points $\pm 1$. Let $B$ be the union of $G$ with the two small disks centered at the points $1, i$. Then, being contractible CW subcomplexes of $S^2$, both $A, B$ are retracts of $S^2$. On the other hand, there is no homeomorphism $(S^2,A)\to (S^2,B)$. Lastly, $A$ is clearly homeomorphic to $B$.

Moishe Kohan
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  • Thank you for your answer: Interesting Idea. May you elaborate why the pairs are not homeomorphic?There is an intuitional reason indicating that two pairs you mentioned are homeomorphic: – Ali Taghavi Aug 17 '16 at 07:25
  • I think that there is a homeomorphism of $\mathbb{R}^{2}$ which send the first quadrant to the upper half plane. sending 1 to 1 and i to -1. Now we blow up these point to obtain discs. – Ali Taghavi Aug 17 '16 at 07:28
  • @AliTaghavi: No, you cannot do it. Of course, you can send a quadrant to the upper half-plane, but try to formalize what do you mean by "blow up". The reason a homeomorphism in my example does not exist is that you cannot change the cyclic order on the link of $0$ in the graph $G$ by an ambient homeomorphism of $S^2$. (Of course, you can do it by a homeomorphism $G\to G$.) And a homeomorphism cannot send a point to a disk. – Moishe Kohan Aug 17 '16 at 07:43