Q:

Euler characteristic of closed orientable surface of dimension $4n + 2$.

I'm considering the Euler characteristic of closed orientable surface of dimension $4n + 2$. The goal is to show that it is even. 
By Poincare Duality, and applying the Universal Coefficient Theorem,
I was able to show that $H_i(M, \mathbb{Z}) \cong H^{4n + 2- i}(M, \mathbb{Z})$. So $\chi(M) = (-1)^{2k + 1}H_{2k + 1}(M, \mathbb{Z})$.
However, I have no information about homology group on this level. It being $2k + 1$ also makes me think that I should be using some odd-ness of the homology group here. 
I searched online and saw that there are some cup product argument but there are some gaps that I couldn't follow. 
Any help will be appreciated! 

A:

It's not true that $\chi(M)=(-1)^{2k+1}
\dim H_{2k+1}(M,\Bbb Z)$, rather
 $\chi(M)\equiv\dim H_{2k+1}(M,\Bbb Z)\pmod2$, but that's all you need.
It's easier to work with real coefficients. Poincare duality
means that the cup product $H^{2k+1}(M,\Bbb R)
\times H^{2k+1}(M,\Bbb R)\to H^{4k+2}(M,\Bbb R)\cong\Bbb R$ is a perfect
pairing. But since $2k+1$ is odd, $\alpha\smile\alpha=0$
for all $\alpha\in H^{2k+1}(M,\Bbb R)$. This cup product is
therefore symplectic, and one cannot have a perfect symplectic pairing
on an odd-dimensional space. Therefore $\dim H^{2k+1}(M,\Bbb R)$
is even, and so is $\chi(M)$.

