Euler number zero for odd dimensional compact manifolds I need to prove that
every compact manifold of odd dimension has Euler number zero.
The Euler number of $M$ compact and oriented is
$$
e(M):=\int_Ms_0^*\phi(TM)
$$ 
where $s_0$ is the zero section of $TM$ and $\phi(TM)$ is its Thom class.
We also proved that
$$
e(M)=\sum_q (-1)^qh_{DR}^q(M)
$$
where $h_{DR}^q(M):=dim H_{DR}^q(M)$ and $H_{DR}^q(M)$ is the $q-th$ cohomology ring of $M$.

Let's suppose $M$ orientable. And let $dim(M)=2n+1$.
By Poincaré duality, we get 
$$
H^q(M)\cong (H^{2n+1-q}(M))^*
$$
for every $q$.
Since every compact manifold is of finite type (hence its cohomology rings are finite dimensional) and since every finite dimensional space is isomorphic to its dual, we get
$$
h^q(M)=h^{2n+1-q}(M)
$$
for every $q$.
Let's apply the formula
$$
e(M)=\sum_q(-1)^qh^q(M)
$$
We see that all the terms $h^q(M)$ and $h^{2n+1-q}(M)$ delete.
