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This is exercise 33 (p.158) from section 2.2 in Hatcher's Algebraic Topology:

Suppose the space $X$ is the union of open sets $A_1, \ldots, A_n$ such that each intersection $A_{i_1} \cap \cdots \cap A_{i_k}$ is either empty or has trivial reduced homology groups. Show that $\tilde{H_{i}}(X) = 0$ for $i \geq n-1$.

I'm pretty sure Mayer-Vietoris needs to be used, and that there should be some induction going on, but I haven't been able to figure it out.

user5826
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user237334
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    Use Mayer-Vietoris one by one on the unions $A_1$, $A_1 \cup A_2$, $(A_1 \cup A_2) \cup A_3$, ... –  May 02 '15 at 21:41
  • Are you sure? I thought of that, but as a conseuqence of that argument, I seem to get that all reduced homology groups of X are trivial. That can't be...unless I'm missing something. – user237334 May 02 '15 at 22:32
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    Certainly Mayer-Vietoris is not wrong. Perhaps you should write out an answer where you try to prove it using that idea? You'll either find your mistake in doing so, or someone else will. –  May 02 '15 at 22:33
  • I imagine what's going wrong is that Mayer-Vietoris does not directly translate to reduced homology groups if you're covering a space $X$ by sets $A$ and $B$ with empty intersection. Consider the case where $X$ is the disjoint union of two discs, then $\tilde{H}_0(X)=\mathbb{Z}$. On the other hand, using Mayer-Vietoris without any tweaks will give you the (incorrect) answer that $\tilde{H}_0(x)\simeq\tilde{H}_0(A)\oplus\tilde{H}_0(B)\simeq 0$. You can, however use the augmented short exact sequence of chain complexes that Hatcher mentions in the Mayer-Vietoris section. – Shehzad Ahmed May 03 '15 at 13:09
  • @ShehzadAhmed Why is $\widetilde{H_0}(D^2 \sqcup D^2) \cong \Bbb Z$? $\widetilde{H_0}$ is reduced homology, right? So $\widetilde{H_0}(D^2 \sqcup D^2) \cong \widetilde{H_0}(pt \sqcup pt) \cong \oplus \widetilde{H_0}(pt) \cong 0$. – Balarka Sen May 03 '15 at 13:13
  • Furthermore, if you use the unreduced Mayer-Vietoris long exact sequence, that gives you $\Bbb Z^2 \cong H_0(D^2 \sqcup D^2) = H_0(X) \cong H_0(A) \oplus H_0(B)$, and the latter is really $\Bbb Z^2$. – Balarka Sen May 03 '15 at 13:22
  • @BalarkaSen, We obtain the reduced homology groups by way of an augmented chain complex $\cdots \to C_2(X)\to C_1(X)\to C_0(X)\to \mathbb{Z}\to 0$. Here, the map from $C_0(X)$ to $\mathbb{Z}$ takes a chain $\sum n_i\sigma_i$ to $\sum n_i$. In this case case for $X=D^2\sqcup D^2$, we have two zero cells and two one cells. Both of those one cells are boundaries, so the map $C_1\to C_0$ is just the zero map. The map $C_0 \to \mathbb{Z}$ has a kernel of $\mathbb{Z}$, and thus $\tilde{H}_0(X)\simeq \mathbb{Z}$. – Shehzad Ahmed May 03 '15 at 13:29
  • @BalarkaSen, the question was about reduced homology groups, and the asker was wondering why using Mayer-Vietoris would give him the incorrect answer. I was simply noting that Mayer-Vietoris needs a small tweak to translate to the case of reduced homology groups. – Shehzad Ahmed May 03 '15 at 13:32
  • @ShehzadAhmed "two zero cells and two one cells"? $C_n(X)$ are formal $\Bbb Z$-linear combination of singular $n$-simplices in $X$, not cells. – Balarka Sen May 03 '15 at 13:33
  • @ShehzadAhmed I get it, but what you wrote above about $\widetilde{H_0}(X) = \Bbb Z$ is not true. – Balarka Sen May 03 '15 at 13:36
  • Let us continue this discussion in chat. – Shehzad Ahmed May 03 '15 at 13:36
  • For the benefit of the op and anyone else that reads this discussion, the reduced zeroth degree homology of a spaces with $n$ path components is $\mathbb{Z}^{n-1}$. This is because $\tilde{H}_0(X)\oplus\mathbb{Z}\cong H_0(X)$ for non-empty $X$, which can be deduced from the augmented singular chain complex and some homological algebra. – Dan Rust May 06 '15 at 13:44
  • @user237334 Using the hint from the first comment, how are you getting all reduced homology groups of $X$ trivial? – user5826 Mar 16 '20 at 00:16

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