Cup product on the torus, $\alpha \cup \alpha \neq 0$? I want to compute the cup product on the Torus. But somehow my calculation is wrong and I cannot figure out why.
We give $T^2$ the $\Delta$-complex structure as in the picture below:
($K,L$ are positively oriented, starting in the bottom left corner)

I computed the cohomology groups with integer coefficients to be
$$H^0=<x^*>, \qquad H^1=<A^*+C^*,B^*+C^* >, \qquad H^2=<K^*>,$$
where $^*$ denotes the dual, e.g. $A^*(A)=1, A^*(B)=0=A^*(C)$.
Due to the anti symmetry of the cup product, we have $$(A^*+C^*) \cup (A^*+C^*)=-(A^*+C^*) \cup (A^*+C^*),$$ so $$(A^*+C^*) \cup (A^*+C^*)=0.$$
But if I compute it explicitly, I get
$$(A^*+C^*) \cup (A^*+C^*) (K)= (A^*+C^*)(C) * (A^*+C^*)(-A)=-1,\\
(A^*+C^*) \cup (A^*+C^*) (L)= (A^*+C^*)(A) * (A^*+C^*)~(B)~~~=~~~0,$$  
so $(A^*+C^*) \cup (A^*+C^*)$ evaluated on the cycle $K+L$ does not vanish.
This is the wrong result, but I have no clue where I made a mistake. I would appreciate some help.

As A.Rod pointed out, there are ways to parametrize $K$ and $L$ in which the computation yields the right result. For example $K,L$ positively oriented, L starting at the bottom left but K starting at the top right.
However, I don't see any reason why this should depend on the parametrization.
 A: As the original answer was never corrected, I will answer the question myself.
My mistake was, that I did not use a valid $\Delta$-complex structure. For some reason it is mentioned, but not really emphasized in Hatcher's book, that such a structure is more than simply drawing some triangles. 
Take a look at the two "$\Delta$-complexes" below. The left one is the one used in the question.

The left structure has two 2-simplices $[v_0, v_1, v_2]$ and $[w_0, w_1, w_2]$. These have the faces $[v_0, v_1],~ [v_1,v_2],~ [v_0,v_2]$ and $[w_0, w_1]~,[w_1, w_2],~ [w_0,w_2]$.
The picture says how these faces have to be identified. In particular, considering $A$, we have to identify $[v_0,v_1]$ with $[w_2, w_1]$. But $[w_2, w_1]$ is not a face! Thus this is not a valid $\Delta$-complex structure.  
The right picture however describes a $\Delta$-complex, as it has the faces $$[v_0,v_1] \sim [w_1, w_2] ~(A)\\
[v_1,v_2] \sim [w_0, w_1] ~(B)\\
[v_0,v_2] \sim [w_0, w_2] ~ (C)$$

Computing the cohomology groups one gets
$$H^0=<x^*>, ~~~ H^1=<A^*+C^*, B^*+C^*> ~~~H^2=<K^*,L^*|K^*-L^*>$$
and $K+L$ is a $2$-cycle.
Further, 
$$(A^*+C^* \cup A^*+C^*)(K) = (A^*+C^*) (B) (A^*+C^*)(A)=0 \\
(A^*+C^* \cup A^*+C^*)(L) =(A^*+C^*)(A) (A^*+C^*)(B) =0$$
So $(A^*+C^* \cup A^*+C^*)(K+L) =0$. Similarly one gets $(A^*+B^* \cup A^*+B^*)(K+L) =0$.
One couldthink that this result only holds, because everything in the calculation above vanishes. However, changing the basis of $H^1$ to $H^1=<A^*-B^*,A^*+C^*>$ still gives the desired result.
A: View your square as (the convex hull of) $s_0=(0,0),s_1=(0,1), s_2=(1,1), s_3=(1,0)$. The orientation on $A=[s_0, s_1]$, $B=[s_1, s_2]$ and $C=[s_0, s_2]$ is given by the order of the vertices and $K$ will be ordered as $[s_0, s_1, s_2]$ and $L$ as $[s_2, s_3, s_0]$.
This way you have $\partial K=B-C+A$ and $\partial L=-B+C-A$, thus $K+L$ is closed.
In the same way $(A^*+C^*)(\partial K)=(A^*+C^*)(\partial L)=0$ and $A^*+C^*$ is closed.
Now $$(A^*+C^*)\cup(A^*+C^*)(K+L)=(A^*+C^*)(A)(A^*+C^*)(B)+(A^*+C^*)(-A)(A^*+C^*)(-B)=0$$
and $$(B^*+C^*)\cup(B^*+C^*)(K+L)=(B^*+C^*)(A)(B^*+C^*)(B)+(B^*+C^*)(-A)(B^*+C^*)(-B)=0$$
Of course I may have made a mistake in my computation.
