Computing Klein bottle's cohomology ring in $\mathbb{Z}$ Well I've been struggling with this one.
This is the picture of the Klein Bottle. It has two triangles (U upper, V lower), three edges (the middle one is "c") and only one vertex repeated 4x.

So my problem essentially is with the relation between the Delta-simplex structure and cohomology.
So suppose that we have the previous structure in the picture of the Klein Bottle. We represent by $T^*$ the dual of $T$. By computing the simplicial cohomology we get that $\Delta^0=\langle v^*\rangle$, $\Delta^1=\langle a^*, b^*, c^*\rangle$, $\Delta^2=\langle U^*, V^*\rangle$.
The coboundary maps satisfy $\delta_0=\delta_1=\delta_3=0$ and $\delta_2(a^*)=U^*-V^*$, $\delta_2(b^*)=-U^*+V^*$ and $\delta_2(c^*)=U^*+V^*$.
So im $\delta_0=$ im $\delta_1=0$, im $\delta_2= \langle U^*-V^* , 2 U^* \rangle$ and $\ker \delta_1=\Delta^0$, $\ker \delta_2=\langle a^*+b^*\rangle$ and $\ker \delta_3=\Delta^2$
As a consequence $H^0=\Delta^0 \cong \mathbb{Z}$ is generated by $v^*$, $H^1 =\langle a^*+b^*\rangle \cong \mathbb{Z}$ and $H^2=\frac{\Delta^2}{\langle U^*-V^* , 2 U^* \rangle}\cong \mathbb{Z}_2$ is generated by both $U^*$ and $V^*$.
The consequence that is not true is the following: $(a^*+b^*)^2=U^*$
This can be proved in this setting quite easily but contradicts the ring structure of the Klein Bottle known as $\mathbb{Z}[x, y]$ modulo the ideal $\langle x^3, 2x^2, xy, y^2 \rangle$
Question: what am I missing?
 A: The issue is that $a^*+b^*$ is not the generator of $H^1$, in fact it is not a cohomology class:
$$(a^*+b^*)(dU) = (a^*+b^*)(a+b-c) = 2.$$
Instead, take $b^*+c^*$ as the generator, which is a cohomology class:
$$(b^*+c^*)(dU) = (b^*+c^*)(a+b-c) = 0,$$
$$(b^*+c^*)(dV) = (b^*+c^*)(a+c-b) = 0.$$
Also, I believe that $\mathbb{Z}[x,y]/(x^3, 2x^2, xy, y^2)$ is not the correct ring structure. For example, the subgroup generated by $1, x, y$ is free of rank 3 but as a group $H^*(K) = \mathbb{Z}^2\oplus\mathbb{Z}/2$.
The integral cohomology ring of the Klein bottle can be described as the free graded ring on generators $|x| = 1$ and $|y| = 2$, modulo degree $\geq 3$ and the relations $x^2 = 0, 2y= 0$. This is the same ring as $\mathbb{Z}[x,y]/(x^2, y^2, xy, 2y)$.
However, if we go to $\mathbb{Z}/2$ coefficients, the square of a degree one element can be the generator of $H^2$. The cohomology ring can be described as a free graded ring on two generators $x, y$ both in degree 1, modulo degree $\geq 3$ and  with relations $xy = yx = 0$, and $x^2 = y^2$. This is the same as $\mathbb{Z}/2[x,y]/(xy, x^2-y^2, x^3,y^3)$. Also, you may find this (and the general case of a genus $g$ non-orientable surface) in Hatcher, Example 3.8, on page 208.
