Homology and cohomology of a manifold So I have this closed, connected 7-manifor $M$ such that $H_1(M; \mathbb{Z})=0$,  $H_2(M; \mathbb{Z})=\mathbb{Z}_2$ and  $H_3(M; \mathbb{Z})=\mathbb{Z}$. I am supposed to calculate $H_i$ and $H^i$ for all $i$. 
Here's my approach. $H_0(M;Z)=\mathbb{Z}$ as it is connected, and from there I use poincare duality to find that $H^7\cong H^4 \cong \mathbb{Z}$ ,$H^6 \cong 0$, and $H^5 \cong \mathbb{Z}_2$. 
Which is all fair and good, but I am stuck on trying to find the other homology and cohomology groups. My first instinct was to use the universal coefficient theorem to find the other cohomologies but that didn't work, any hints would be appreciated, thank you!!!
 A: To get $H^4$, we don't need to use the UCT (as suggested by your comment). It's better to use Poincare duality (as suggested by your original post):
$$ H^4(M, \mathbb Z) \cong H_3(M, \mathbb Z) \cong \mathbb Z.$$
Similarly,
$$ H^5(M, \mathbb Z) \cong H_2(M, \mathbb Z) \cong \mathbb Z_2,$$
and likewise for $H^6$ and $H^7$. 
The trickiest part of this exercise is getting $H_4$, $H_5$, $H_6$ and $H_7$. To do this, we need to do some reverse engineering with the Universal Coefficient Theorem.
First, let us remind ourselves of what the relevant ${\rm Hom}$ and ${\rm Ext}$ groups look like:
$$ {\rm Hom}(\mathbb Z, \mathbb Z) = \mathbb Z, \ \ \ \ \ \ \ \ {\rm Ext}(\mathbb Z, \mathbb Z) = 0.$$
$$ {\rm Hom}(\mathbb Z_2, \mathbb Z) = 0, \ \ \ \ {\rm Ext}(\mathbb Z_2, \mathbb Z) = \mathbb Z_2.$$
In view of this, we can rephrase the UCT as follows:

$H^i$ is isomorphic to the direct sum of all the $\mathbb Z$ components from $H_i$ and all the $\mathbb Z_2$ components from $H_{i-1}$.  

This statement tells us how to get cohomology from homology. We need to reverse-engineer it to get homology from cohomology:

$H_i$ has as many $\mathbb Z$ components as $H^i$, and as many $\mathbb Z_2$ components as $H^{i+1}$.

So given that $H^4$ has one $\mathbb Z$ component and $H^5$ has one $\mathbb Z_2$ component, it must be the case that
$$ H_4(M, \mathbb Z) \cong \mathbb Z \oplus \mathbb Z_2.$$
Similar logic will get us $H_5$, $H_6$ and $H_7$.
