TITLE: Simplicial cup product on torus
QUESTION [6 upvotes]: I'm trying to compute the simplicial cup product on the torus (using $\Delta$-complexes) but running into a problem: each way I draw the fundamental polygon I get different answers!  When I draw it as Hatcher does in 2.1, and label the vertices as on the left below,
         
(original image)
I get the correct answer, but I get the wrong answer when I use the triangulation on the right (or any other ones I tried).  Why is this happening?  What's special about Hatcher's triangulation?
Edit: Let $\alpha$ and $\beta$ generate $H^1(X; \mathbf{Z})$, dual to $a$ and $b$, respectively, and let $\varphi$, $\psi$ be representative cochains. Then using the triangulation on the right, I get $$\alpha \smallsmile\alpha(A) = \varphi(A|[01])\varphi(A|[13]) = \varphi(a)\varphi(c) = 1$$ where $A$ is the upper 2-simplex, and this is already wrong.

REPLY [2 votes]: The problem is that you are performing an operation which is "not allowed": The cup product $\alpha\smile\alpha$ is a cohomology class in $H^2(X;\mathbb Z)$. Now such a class is determined only up to coboundaries: If $φ$ and $\psi$ represent the same class, then $φ-\psi$ is a coboundary, and that means it vanishes on cycles. So if $C$ is a cycle in $C_2(X)$, then $φ(C)=\psi(C)$. We cannot expect $φ$ and $\psi$ to agree on ordinary simplices, like the upper triangle $A$ in the right-hand diagram. So we should not apply a cohomology class to an ordinary chain, but only to cycles, or to homology classes.
On the cycle $A-B$, $α⌣α$ gives $0$, since $α(A)=α(B)=-1$ (I'm taking $-1$ instead of $1$ since in my calculation the cocycle $φ$ representing $\alpha$ maps $a$ to $1$, $c$ to $-1$, and $b$ to $0$, and I think a $φ$ which sends $c$ to $1$ only works if $c$ points in the opposite direction).