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I came across this question. Prove that $\tilde{H}_i(S^n-X)\cong H_{n-i-1}(X)$ if $X$ is a finite connected graph embedded in $S^n$. By Alexander Duality, this is true if the group on the right is a cohomology group instead of homology. I've been trying to use excision, possibly followed by the long exact sequence of the pair, but I can't seem to find the correct subspaces.

EDIT: This seems to be false. If you let $S^1\subseteq S^2$ be the equator, then $S^2-S^1\simeq S^0$. Then $\tilde{H}_1(S^0)=0$, by $H_{2-1-1}(S^1)\cong \mathbb{Z}$. So, the problem has a typo? Is there a way to adjust it and make it true?

J126
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1 Answers1

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The right was supposed to be reduced. That at least gets rid of the problem in your counterexample.

Adding the reduced to the right and moving the lower index to an upper index on the right is essentially just Alexander duality as you mention. You also need to take an open tubular neighbourhood of $X$ so that the complement is compact, but this doesn't affect anything up to homotopy. The universal coefficient theorem tells you that $H_k(Y)=H^k(Y)$ as long as the homology of $Y$ has no torsion (which is the case when $Y$ is a graph) so you get the (modified) result.

Specifically $$\tilde{H}_i(S^n-X) \cong \tilde{H}_i(S^n\setminus u(X)) \cong\tilde{H}^{n-i-1}(u(X)) \cong \tilde{H}^{n-i-1}(X)\cong \tilde{H}_{n-i-1}(X)$$

where $u(-)$ means 'take an open tubular neighbourhood'.

Dan Rust
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  • Yes. But, is there an isomorphism if the right hand side is reduced homology? – J126 Jul 11 '14 at 14:44
  • You mean when using the universal coefficient theorem? This just reduces the rank of the $0$th (co)homology by one so yes it is still an isomorphism. – Dan Rust Jul 11 '14 at 14:49
  • I found this problem on a qualifying exam. The exam assumes no knowledge of cohomology. That's why I tried excision. Do you know of a proof without all the technology of cohomology? – J126 Jul 11 '14 at 14:54
  • If you want to not use alexander duality, then probably using mayer-vietoris and considering a ball around the graph as one of the subsets, and then the other set being the sphere with the ball removed will work. I'm afraid I have to go now so I don't have the time to check how this would work, but feel free to try it. – Dan Rust Jul 11 '14 at 14:58
  • Just remember than an $n$-ball with a subgraph removed will be homotopy equivalent to an $n-1$ sphere with a certain number of circles wedged onto it which is determined by the rank of the first homology of the graph (equivalently the number of circles the graph is homotopy equivalent to the wedge of). – Dan Rust Jul 11 '14 at 15:00