# Problem of the Week #282 - Nov 20, 2018

Status
Not open for further replies.

#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

-----
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.

-----

#### Euge

##### MHB Global Moderator
Staff member
Here is a hint: Consider using Poincaré duality and the universal coefficients theorem.

#### Euge

##### MHB Global Moderator
Staff member
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.

Status
Not open for further replies.