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I have to solve the following questions:

"For a subset $X \subset S^n$ determine the homology group $H_i(S^n - X)$, where
(a) $X \cong S^l \vee S^k$
(b) $X \cong S^l \sqcup S^k$ (disjoint union) "

The second one I could solve by myself: I used de Morgan's rule, the Proposition 2B.1. from Hatcher's Algebraic topology (Brouwer, Jordan) and a long exact Mayer-Vietoris sequence (if someone wants to see it she/he may ask for it). But I'm not able to see what I have to do in the first case (a)...could perhaps someone give me a hint? Thank you very much!

Donut
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    I think there is a typo, you mean to say "wedge $\vee$" not "smash product $\wedge$"... am I right? otherwise, in case of smash product there may not exists any embedding of $X$ is $S^n$ – Anubhav Mukherjee Jan 03 '16 at 14:00
  • Oh yes!!! You're right! I have edited it. – Donut Jan 03 '16 at 15:10
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    The only way I know to pass from a (homeomorphism type) of subset and learn about the homology of its complement, where I'm not given the actual subset (just what it's homeomorphic to), is Alexander duality and things of its ilk. –  Jan 03 '16 at 15:30
  • I have looked it up, but we didn't speak in class about this Alexander duality at all. There has to be a rather basic possibility. Perhaps one can use excision...I try it out. –  Jan 03 '16 at 17:52

1 Answers1

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The argument you've outlined for (b) works for (a) after a shift in perspective:

Suppose $L$ and $K$ are closed subspaces of $S^n$. Let $A=S^n- L$ and $B=S^n -K$. Then we have $A \cap B=S^n -(L \cup K)$, and the Mayer-Vietoris sequence reads $$H_{i+1}(A \cup B)\to H_i(S^n-(L\cup K)) \to H_i(S^n - L) \oplus H_i(S^n-K) \to H_i(A \cup B).$$ In part (b), you had $L=S^l$ and $K=S^k$ disjoint. This means $A \cup B=S^n$, so you get isomorphisms in the middle of the above sequence for all $i \leq n-2$. But in part (a), $L=S^l$ and $K=S^k$ intersect at a single point -- the wedge point. So we have $$A \cup B = (S^n - L) \cup (S^n - K) = S^n - (L \cap K)=S^n -\{pt\} \cong \mathbb{R}^n.$$

Leveraging the boring homology of $\mathbb{R}^n$ and the exactness above, we get isomorphisms $$H_i(S^n-(L \cup K)) \cong H_i(S^n - L) \oplus H_i(S^n - K)$$ for all $i\geq 0$. (The $i=0$ case can be argued using reduced homology, for example.) Now use the result from Hatcher again, just as in part (b).

Kyle
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  • Thank you so much for this nice proof. It was indeed the same thing as in (b). I just did not see the correct Mayer-Vietoris Sequence! – Donut Jan 05 '16 at 10:41