Cup product on torus I want to calculate the cup product on torus (cf. Hatcher's book)
If $\pi_1(T^2) = ([a]) + ([b])$, then by universal coefficient theorem we have 
a cocycle $\alpha$ (resp. $\beta$) which have a value $1$ only on a loop $a$ (resp. $b$). 
Let $\sigma$ is a $2$ cycle with $([\sigma])=H_2(T^2)$ so that $\partial \sigma = a+ b - a-b$. Here in $\Delta$-complex notation, $\sigma$ is a square with four edges $a$, $b$, $-a$, $-b$ 
Hence $$\alpha \cup \beta (\sigma) = \alpha (a) \beta (b-a-b) = 1\cdot 0 =0 $$
But on the other hand, $$0\neq \alpha \cup \beta (\sigma) = \alpha (a+b) \beta (-a-b) = 1\cdot (-1) = \beta (a+b) \alpha (-a-b) =  \beta\cup \alpha (\sigma)$$ so that $$
      \alpha \cup \beta (\sigma) \neq - \beta\cup \alpha (\sigma)$$
What is wrong ? 
 This calculation is based on the definition of cup product :
 $$  \phi\cup\psi (\sigma ) = \phi( \sigma|_{[v_0, .... , v_k]})\psi (\sigma|_{[v_k, ... ,
v_{k+l}]} ) $$. So on $T^2$, 
  by taking $\sigma|_{[v_0v_1]} = a$ and $\sigma|_{[v_1v_2]} = b-a-b$ or 
  $\sigma|_{[v_0v_1]} = a+b$ and $\sigma|_{[v_1v_2]} = -a-b$, we have the above calculation.
Please give me a some advice or correction. Thank you in advance. 
 A: When you write $\alpha \cup \beta (\sigma) = \alpha (a+b) \beta (-a-b)$, I think you may be confused about how the cup product acts on singular chains.  The "front face back face" rule is applied to each term in the formal sum separately and then these products are added together afterward. I.e. 
$\alpha \cup \beta (\Sigma _i c_i \sigma _i ) = \Sigma _i c_i \alpha (\sigma _i|_{[v_0,v_1]} )\beta (\sigma _i|_{[v_1,v_2]})$ 
not
$\alpha (\Sigma _i c_i \sigma _i|_{[v_0,v_1]})\beta (\Sigma _i c_i \sigma _i|_{[v_1,v_2]})$
A: See the following page : 
http://math.ucsd.edu/~bewilson/qualprep/TorusCupProductCalculation.pdf 
Consider a square $ABCD$ which is a union of triangle $ABC$ and triangle $ACD$.
Here, it contains 5 edges and represents $T^2$. 
Let $\alpha $ (resp. $\beta$) be a $1$-cocylce which values 1 only on $AB$, $DC$, $AC$ 
  (resp. $BC$, $AD$, $AC$). 
Then $$ \alpha \cup \beta ( ABCD) =  \alpha \cup \beta (ABC  + ACD) = 
  \alpha (AB)\beta (BC) + \alpha (AC)\beta(CD) = 1 +0 =1$$
And $$ \beta \cup \alpha ( ABCD) =  \beta \cup \alpha (ABC  + ACD) = 
  \beta (AB)\alpha (BC) + \beta (AC)\alpha(CD) = 0 + 1(-1) =-1$$
