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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. – 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
To make all this explicit, try looking at the case when $X$ is just a point. – Chris Eagle May 14 '12 at 11:13
@hgbreton: M-V does not tell you nothing there. Write it out and see. – Chris Eagle May 14 '12 at 18:02

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