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Problem of the Week #282 - Nov 20, 2018

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Suppose $X$ is a closed connected orientable manifold of dimension $2n$. Prove that if the homology group $H_{n-1}(X)$ is torsion free, then $H_n(X)$ is also torsion free.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is a hint: Consider using Poincaré duality and the universal coefficients theorem.
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
No one answered this week's problem. You can read my solution below.

By Poincaré duality and the universal coefficients theorem, it follows that
$$H_n(X) \approx H^n(X) \approx H_n(X)^{\text{free}} \oplus H_{n-1}(X)^{\text{torsion}}$$
Since $H_{n-1}(X)$ is torsion-free, then $H_{n-1}(X)^{\text{torsion}} = 0$. Therefore $H_n(X) \approx H_n(X)^{\text{free}}$, showing that $H_n(X)$ is also torsion-free.
 
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