Intersection form on quotient manifold Let $E_{1},E_{2}$ be elliptic curves over $\mathbb{C}$. We denote by $\iota_{i}$ the translation by a 2-torsion point on $E_{i}$. Then $G=\mathbb{Z}/2\mathbb{Z}$ acts freely on the the product $E_{1}\times E_{2}$ via the involution $\iota=(\iota_{1},\iota_{2})$. and the quotient 
$$
X=(E_{1}\times E_{2})/G
$$
is a 4-dimensional manifold (complex surface). I would like to understand the intersection form on the middle cohomology
$$
(-,-)_{X}:H^2(X,\mathbb{Z})\times H^2(X,\mathbb{Z}) \rightarrow H^4(X,\mathbb{Z})\cong \mathbb{Z}
$$
via the cup product. I initially thought

There is a ono-to-one correspondence
  between  $$ H^{2}(X,\mathbb{Z})
> \longleftrightarrow H^{2}(E_{1}\times
> E_{2},\mathbb{Z})^{G}, $$ Since the
  action is free, the intersection form
  on $H^{2}(X,\mathbb{Z})$ is given by
  the intersection form on
  $H^{2}(M\times N,\mathbb{Z})^{G}$
  divided by $|G|$. So, any intersection
  number on $H^{2}(M\times
> N,\mathbb{Z})^{G}$ must be a multiple
  of $|G|=2$.
On the other hand, we have  $$
> p_{1}^{*}(\alpha_{E_{1}}), \
> p_{2}^{*}(\alpha_{E_{2}})\in
> H^{2}(E_{1}\times
> E_{2},\mathbb{Z})^{G} $$ (because $G$
  preserves both $E_{1}$ and $E_{2}$)
  and  $$ p_{1}^{*}(\alpha_{E_{1}})\cup
> \
> p_{2}^{*}(\alpha_{E_{2}})=\alpha_{E_{1}\times E_{2}} $$ where
  $H^{\dim_{\mathbb{R}}(M)}(M,\mathbb{Z})\cong
> \mathbb{Z}\alpha_{M}$ via the natural
  orientation and $p_{i}$ is the $i$-th
  projection of $E_{1}\times E_{2}$.
  This means that the intersection
  number $p_{1}^{*}(\alpha_{E_{1}})\cup \
> p_{2}^{*}(\alpha_{E_{2}})$ is 1, not
  divisible by $|G|=2$.

When I asked a similar question, some people pointed out that the correspondence 
$$
H^{2}(X,\mathbb{Z}) \ \longleftrightarrow H^{2}(M\times N,\mathbb{Z})^{G},
$$
does not hold in general; there is the Hochschild-Serre spectral sequence
$$
E^{p,q}=H^{p}(G,H^{q}(E_{1}\times E_{2},\mathbb{Z}))\Rightarrow H^{p+q}(X,\mathbb{Z})
$$
Here $E^{0,2}$ term corresponds to $H^{2}(M\times N,\mathbb{Z})^{G}$ above.  
Having said that, I still don't quite understand the intersection form on $X$ (mainly due to my poor understanding of the Spectral sequence). I would appreciate it if anyone could describe the intersection form. What if $\iota_{2}$ is replaced by $-id_{E_{2}}$? 
 A: Luckily, the $X$ thus described is a torus. This means that $H^2(X,\mathbb Z)$ has a nice explicit description: It is $\wedge^2 H^1(X,\mathbb Z)$. The intersection form is the the symmetric bilinear map to $\wedge^4 H^1(X,\mathbb Z)=\mathbb Z$.  $H^1(X,\mathbb Z)$ is an index two sublattice of $H^1(E_1 \times E_2,\mathbb Z)$ , which I guess you can see from the exact sequence for homotopy groups of a fibration.
In fact, the action of $G$ on the cohomology groups is trivial, because it is homotopic to the identity, since the group of translations is connected!
So $H^2(X,\mathbb Z)$ lies in $H^2(E_1 \times E_2,\mathbb Z)$ and similarly $H^4(X,\mathbb Z)$ lies in $H^4(X,E_1 \times E_2)$, so you do indeed get a division by $|G|$ when you identify the $H^4$s with $\mathbb Z$. But this does not mean that all intersection forms in $H^2(E_1\times E_2,\mathbb Z)$ are even, just those for cocycles which are pullbacks of cocycles from $X$. The criterion for this is not $G$-invariance of the cohomology, as the $G$ action on the cohomology is trivial, it's $G$-invariance of the underlying cocycle.
