Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If two topological spaces have the same homotopy type, how do I prove that their complements in an ambient space are homeomorphic?

share|improve this question
1  
Well my specific example was that I have a 2-complex which is homotopy equivalent to the 2 sphere. This directly implies that the its complement in the 3-sphere is the union of two open 3-balls. How do I show that this is true? –  user7485 Feb 22 '11 at 9:41

3 Answers 3

I don't know how your question can be fixed so as to get a true statement, but as asked this is simply wrong. Take the $x$-axis in $\mathbb{R}^{2}$. It is contractible and its complement has two connected components. However, the complement of a point only has one connected component, so this is an easy example of two subspaces of the same homotopy type whose complements aren't homeorphic or even homotopic.

share|improve this answer
    
Well my specific example was that I have a 2-complex which is homotopy equivalent to the 2 sphere. This directly implies that the its complement in the 3-sphere is the union of two open 3-balls. How do I show that this is true? –  user7485 Feb 22 '11 at 10:23
    
@346rte: I'm not an expert in algebraic topology, so you have to be a bit patient until one of those comes along (e.g. @Ryan Budney). Who tells you that "this directly implies that..."? I guess that the hypothesis is strong enough to exclude stuff like the horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere but I'm not sure. Your question should follow from the $h$-cobordism theorem en.wikipedia.org/wiki/H-cobordism, but that is probably using a sledgehammer to crack a nut. –  t.b. Feb 22 '11 at 10:43
1  
346rte: I can't see that page on Google Books, and my copy must be at the office. I suggest you edit your question to specifically explain the situation you're interested in. –  Dan Ramras Feb 23 '11 at 2:49

Indeed, the whole subject of knot theory is interesting because different subsets of $R^3$ that are homeomorphic to each other (namely different knots, each homeomorphic to a circle) have non-homeomorphic complements in $R^3$. And knots are homeomorphic, not just homotopy equivalent.

share|improve this answer

Alexander duality is the closest to what you seem to want: http://en.wikipedia.org/wiki/Alexander_duality.

share|improve this answer

Your Answer

 
discard

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