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Given that $H_n(X)$ is free abelian, I'm trying to find the homology groups of $Y=S^1\times X$ using the Mayer-Vietoris theorem.

My first attempt decomposed $Y$ as $A \cup B$ where $A=\{*\}\times X$ and $B = S^1 \setminus \{*\}\times X$. Then $B$ clearly homotopy equivalent to $A$ and the Mayer-Vietoris sequence gives that $H_n(Y)=H_n(X)\bigoplus H_n(X)$.

This is clearly false (the torus is one easy counterexample). I assume that where I went wrong in is assuming that $B$ was triangulable. Am I right?

What would be the right way to go about this? Apart from what I tried I see no other decompositions which would work! I must be missing something.

Anon
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Edward Hughes
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    Start by reading the statement of Mayer-Vietoris you are using. The usual form requires that the interiors of $A$ and $B$ cover $Y$. This is not the case here. – Chris Eagle May 14 '12 at 10:23
  • Ah that's a subtlety I hadn't picked up on! So if I modify my $A$ slightly so $A=U\times X$ for some small open set $U$ in $S^1$ then I should be able to use Mayer-Vietoris, right? In that case then $A\cap B=S^1$ and it's starting to look more correct! Thanks. – Edward Hughes May 14 '12 at 11:07
  • No, $A \cap B$ is homeomorphic to $((-1,0) \cup (0,1)) \times X$, and is homotopy equivalent to the disjoint union of two copies of $X$. – Chris Eagle May 14 '12 at 11:12
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    To make all this explicit, try looking at the case when $X$ is just a point. – Chris Eagle May 14 '12 at 11:13
  • Ah yes of course. That makes more sense. Can't see immediately that Mayer-Vietoris is now going to be useful though... – Edward Hughes May 14 '12 at 11:23
  • Is it required to use Mayer-Vietoris? If you have covered the Kunneth formula, this should come out pretty easily – Juan S May 14 '12 at 11:23
  • I've read about the Kunneth formula and can see that it would be much easier using that, but the problem I'm trying to do explicitly states that I should use Mayer-Vietoris! – Edward Hughes May 14 '12 at 11:24
  • Think about how you'd go about using Mayer-Vietoris to compute homology of $S^1$ -- what open sets $U, V$ you'd use. Then use $U \times X$ and $V \times X$. – Thomas Belulovich May 14 '12 at 13:21
  • Surely I'd just use any two overlapping $U$ and $V$ with interiors covering $S^1$? Then both of these would be contractible, and the Mayer-Vietoris sequence tells us nothing... Could you give me a bigger hint? – Edward Hughes May 14 '12 at 13:31
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    @hgbreton: M-V does not tell you nothing there. Write it out and see. – Chris Eagle May 14 '12 at 18:02
  • Write $Y=S^1\times X$ Okay so we have exact sequence $\dots\rightarrow H_n(Y)\rightarrow H_n(X) \bigoplus H_n(X)\rightarrow H_n(X) \bigoplus H_n(X) \rightarrow H_{n-1}(Y)\rightarrow \dots$

    This doesn't seem very useful to me...

     – Edward Hughes May 14 '12 at 19:21
  • The real question is: what are the maps? – Aaron Mazel-Gee May 15 '12 at 01:17
  • Writing from left to right the maps are $\delta_, \alpha_, \beta_$. I presume that I'm meant to say something about $\alpha_$ in particular right? Am I correct in thinking that it is the zero map for $n\geq 1$? And the identity map for $n=0$? – Edward Hughes May 15 '12 at 12:05
  • @hgbreton: First of all, I think you have the signature wrong: Given $X=A\cup B$, all of the "geometric" maps -- which are the degree-preserving ones -- are induced by inclusions. That is, you should have $H_n(A\cap B) \rightarrow H_n(A) \oplus H_n(B) \rightarrow H_n(X)$. There are two possible conventions for these maps: writing $i_U^V$ for the inclusion of $U$ into $V$, you can either take the first map to be $((i_{A\cap B}^A)*,(i{A\cap B}^B)*)$ and the second map to be $\pm ((i_A^X)* - (i_B^X)_*)$, or... – Aaron Mazel-Gee May 15 '12 at 15:22
  • you can take the first map to be $(\pm (i_{A\cap B}^A)*,\mp (i{A\cap B}^B)*)$ and the second map to be $\pm ((i_A^X)* + (i_B^X)_*)$. I think. Check your book to see what conventions are chosen there. – Aaron Mazel-Gee May 15 '12 at 15:23
  • I was working with the first convention. Is what I said wrong then? I don't really know how to work out $((i^A_{A∩B})∗,(i^B{A∩B})_∗)$ in this context. Could you point me in the right direction? Thanks a lot for your help! – Edward Hughes May 15 '12 at 15:31
  • @hgbreton: The map to study is $H_n((U\cap V) \times X)$ $ \cong H_n((U\cap V)1\times X)\oplus H_n((U\cap V)_2 \times X)$ $ \cong H_n(X) \oplus H_n(X) $ $\rightarrow H_n(U \times X) \oplus H_n(V\times X)$ $\cong H_n(X) \oplus H_n(X)$. (The subscripts denote the two contractible components.) With this convention, the map is $(\alpha,\beta) \mapsto (\alpha + \beta,\alpha + \beta)$. Given that you know $H(X)$ is free abelian, you can determine the kernel and cokernel, and hence $H_(X \times S^1)$ (there will be no extension problems, as you can (and must) show). – Aaron Mazel-Gee May 16 '12 at 08:50
  • Thanks - I see more how to do it now. Just one final question - what is your reasoning for showing that the map is $(\alpha,\beta)\rightarrow(\alpha+\beta ,\alpha+\beta)$? I don't really see why that is true. – Edward Hughes May 16 '12 at 09:43
  • @hgbreton: This is simply because you chose the first convention! Then, for any contractible space $C$, the map $H_n(C \times X) \rightarrow H_n(X)$ is an isomorphism. I'm using this to explicitly identify all the copies of $H_n(X)$ running around (i.e. taking $C \in {(U\cap V)_1,(U\cap V)_2,U,V}$). – Aaron Mazel-Gee May 17 '12 at 04:23
  • Thanks - I understand it now! – Edward Hughes May 17 '12 at 09:36
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    @EdwardHughes Would you like to share your understanding of this problem with others, by posting a solution? As is, the question is listed as Unanswered, which is why I came across it. – ˈjuː.zɚ79365 Jun 17 '13 at 12:03

1 Answers1

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Your choice of $A, B$ does not satisfy the standard prerequisite (the interiors of $A, B$ have to cover $S^1 \times X$) for applying the Mayer-Vietoris sequence. There are also other variants of M-V, but certainly no variant working for $A \cap B = \emptyset$ unless both $A, B$ are open.

See Homology groups of $X \times S^1$ how to do it properly.

Remark.

I intended to close your question as a duplicate, but I think this would not answer where you went wrong.

Paul Frost
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