Problem from Hatcher's book (3.3.25) I have begun to read in Hatcher's book "Algebraic topology", about cohomology. In doing so, I have tried to solve some problems. I have difficulties with problem 3.3.25: Show that if a closed orientable manifold $M$ of dimension $2k$ has $H_{k-1}(M,\mathbb{Z})$ torsion free, then $H_{k}(M,\mathbb{Z})$ is torsion free.
I have no idea. I think that I have to use somehow the Poincare duality, but I don't know how? Can somebody tell me how this works or at least give me a good hint? Thanks in advance. 
mika
 A: The homology of a closed manifold is a finitely generated abelian group, so we can write
$$H_i(M; \mathbb{Z}) \cong \mathbb{Z}^{b_i(M)} \oplus T_i,$$
where $b_i(M)$ is the $i^\text{th}$ Betti number of $M$ and $T_i$ is the torsion part of $H_i(M;\mathbb{Z})$. Then we have that
$$\mathrm{Hom}(H_i(M;\mathbb{Z}), \mathbb{Z}) \cong \mathrm{Hom}(\mathbb{Z}^{b_i(M)},\mathbb{Z}) \oplus \mathrm{Hom}(T_i,\mathbb{Z}) \cong \mathbb{Z}^{b_i(M)}$$
and
$$\mathrm{Ext}(H_i(M;\mathbb{Z}),\mathbb{Z}) \cong \mathrm{Ext}(\mathbb{Z}^{b_i(M)},\mathbb{Z}) \oplus \mathrm{Ext}(T_i,\mathbb{Z}) \cong T_i,$$
where we have used the properties of $\mathrm{Ext}$ found at the bottom of page 195 in Hatcher.
Then the Universal Coefficient Theorem for cohomology says that
$$0 \longrightarrow \mathrm{Ext}(H_{i-1}(M;\mathbb{Z}),\mathbb{Z}) \longrightarrow H^i(M;\mathbb{Z}) \longrightarrow \mathrm{Hom}(H_i(M;\mathbb{Z}),\mathbb{Z}) \longrightarrow 0,$$
or, by the above work,
$$0 \longrightarrow T_{i-1} \longrightarrow H^i(M;\mathbb{Z}) \longrightarrow \mathbb{Z}^{b_i(M)} \longrightarrow 0$$
is (unnaturally) split exact. Hence
$$H^i(M;\mathbb{Z}) \cong \mathbb{Z}^{b_i(M)} \oplus T_{i-1}.$$
Now in particular, since $M$ is a closed, orientable $2k$-dimensional manifold, by Poincaré duality,
$$H^k(M;\mathbb{Z}) \cong H_k(M;\mathbb{Z}),$$
so from the above it must be that
$$T_{k-1} \cong T_k.$$
So if $H_{k-1}(M;\mathbb{Z})$ is torsion-free, i.e. $T_{k-1} \cong 0$, then $H_k(M;\mathbb{Z})$ is torsion-free as well.
