Q:

$\tilde{H}_i(S^n-X)$, $X$ a Finite Graph

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?

A:

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'.

