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Let $(H_*, \partial_*)$ be a homology theory satisfying the dimension axiom. Let $A \subset S^n$ be a proper subset. Show that $H_n(S^n, A)$ is not trivial.

I tried applying the long exact sequence with no success, since $A$ could be virtually anything. I assume $H_n(A) \to H_n(S^n)$ is not injective either, because then I'd get a short exact sequence, but it can not split.

Anything obvious that I missed?

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    $\begingroup$ You want to say that for nice subspaces $A$, that $H_n(A)=0$. (E.g. a subcomplex.) However this is false for general subspaces. If $A$ is a higher dimensional analogue of the Hawaiian Earring space, then Barratt and Milnor showed it has nontrivial singular homology in infinitely many dimensions. So this looks like a tricky problem. What is the source of this problem? $\endgroup$ – Cheerful Parsnip Aug 17 '15 at 18:06
  • $\begingroup$ It's an old homework problem. $\endgroup$ – Cosmare Aug 17 '15 at 18:09
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Okay, it actually is pretty simple. We want to argue that the map $H_k(A)\to H_k(S^n)$ is trivial. This follows because it factors through the contractible space $S^n\setminus\{p\}$ for any point $p\notin A$.

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  • $\begingroup$ Then I get a short exact sequence and I can argue since $H_n(S^n)$ is non trivial, neither can $H_n(S^n,A)$ be? $\endgroup$ – Cosmare Aug 17 '15 at 18:27
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    $\begingroup$ @Cosmare: Yes, because the map $H_n(S^n) \to H_n(S^n,A)$ is injective. $\endgroup$ – user98602 Aug 17 '15 at 18:28

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