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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.

Many thanks in advance!

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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.2012-05-14
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    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.2012-05-14
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    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 $X$.2012-05-14
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    To make all this explicit, try looking at the case when $X$ is just a point.2012-05-14
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    Ah yes of course. That makes more sense. Can't see immediately that Mayer-Vietoris is now going to be useful though...2012-05-14
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    Is it required to use Mayer-Vietoris? If you have covered the Kunneth formula, this should come out pretty easily2012-05-14
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    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!2012-05-14
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    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 $V \times X$.2012-05-14
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    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?2012-05-14
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    @hgbreton: M-V does not tell you nothing there. Write it out and see.2012-05-14
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    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...2012-05-14
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    The real question is: what are the maps?2012-05-15
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    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 $n=0$?2012-05-15
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    @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...2012-05-15
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    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.2012-05-15
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    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!2012-05-15
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    @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).2012-05-16
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    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.2012-05-16
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    @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 $C \in \{(U\cap V)_1,(U\cap V)_2,U,V\}$).2012-05-17
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    Thanks - I understand it now!2012-05-17
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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.2013-06-17

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