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I'm trying to solve the following problem:

Give an example of $Y = A \cup B\cup C$ such that $A,B,C$ are open subsets of $Y$ and the reduced homology groups of $A,B,C, A\cap B, A\cap C, B \cap C, A \cap B \cap C$ are all trivial, the sets $A \cap B, A \cap C , B \cap C$ are non-empty but $H_1 (Y)$ is not trivial. Show that $H_n (Y)$ has to be trivial for $n \geqslant 2$.

It feels like the second part should follow from some exact sequence, but I've tried Mayer-Vietoris to no avail (it's likely i'm missing something though?). As for the example I've tried different partitions of $S^1$ or the torus but I couldn't get it to work. I don't think I have a good enough understanding of what trivial homology groups imply geometrically to come up with an example, other than randomly stumbling onto one. Any help's appreciated.

NewUserForNow
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    Hint for the first part: $A \cap B \cap C$ is allowed to be empty. Try three connected open segments that cover a circle. (Geometrically, trivial homology groups imply either that the space is contractible or that it has been very cunningly constructed - google for "acyclic spaces". Contractible will do for this exercise.) For the second part: have a look at https://math.stackexchange.com/questions/1263699/reduced-homology-groups-of-a-space-which-is-the-union-of-finitely-many-open-subs – Rob Arthan Aug 30 '17 at 21:18
  • @RobArthan Thanks, that example is actually really simple! I read the topic you linked and I'm trying to apply Mayer-Vietoris to $(A \cup B) \cap C$ but i get: $0 \rightarrow H_1 ( (A \cup B) \cap C) \rightarrow H_1 (A \cup B) \oplus H_1 (C) \rightarrow H_1(Y) \rightarrow 0$ and I think the all the groups in the chain other than $H_1(Y)$ are zero which would imply it is too. Where am I wrong? – NewUserForNow Aug 30 '17 at 22:12
  • @RobArthan True, so I prove that $H_n ((A \cup B) \cap C) = H_n ((A \cap C) \cup (B \cap C))$ for $n > 0 $ by Mayer Vietoris, but I cannot say that for $n = 0 $ and same goes for $H_n (Y)$ right? The reason i cannot say that for $H_0$ is because the last groups in the Mayer-Vietoris sequence are a little different for reduced homologies? – NewUserForNow Aug 30 '17 at 22:31
  • @RobArthan Sorry in the first equation i meant to write that they are both equal zero (for n > 0). Yes I see what the problem with $H_1$ is now. If you want to write a short answer based on these comments I will be happy to accept it. – NewUserForNow Aug 30 '17 at 23:05
  • I've left the comment about acyclic spaces as it may be of interest to other passers by, but I'll delete the rest. Feel free to delete your comments or leave them as you wish. – Rob Arthan Aug 30 '17 at 23:48

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For the first part, take $Y$ to be the circle $S^1$ and $A$, $B$ and $C$ to be three connected open segments that cover the circle without any unnecessary overlaps, e.g., $(0^{\circ}, 121^{\circ})$, $(120^{\circ}, 241^{\circ})$ and $(240^{\circ}, 1^{\circ})$. Then $A$, $B$, $C$, $A \cap B$, $A \cap C$ and $B \cap C$ are all contractible and $A \cap B \cap C$ is empty, so these spaces all have trivial reduced homology groups. However the circle $Y = S^1 = A \cup B \cup C$ has $\widetilde{H}_1(Y) = \Bbb{Z}$. This non-trivial $\widetilde{H}_1$ sneaks in because when we look at the Mayer-Vietoris sequence for $Y$ viewed as the union of $A \cup B$ and $C$, we find that $\widetilde{H}_1(Y) \cong \widetilde{H}_0((A \cup B) \cap C)$ and the latter group is $\Bbb{Z}$ (because $(A \cup B) \cap C$ has two connected components).

For the second part, use the Mayer-Vietoris sequence first for the union $A \cup B$ and then for the union $Y = A \cup B \cup C$. See Reduced homology groups of a space which is the union of finitely many open subsets for a generalisation.

Rob Arthan
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