Cohomology ring of the double torus (genus two surface) I computed the cohomology ring of the double torus, i.e. the connected sum $X := T_2\#T_2$, where $T_2 = \mathbb S^1 \times \mathbb S^1$.
I am looking for a way to confirm that my computation looks correct, since I feel I got a weird answer.
I used the fundamental polygon shown in this post, to the left, with $I, II, III, IV$ named $a, b, c, d$ respectively.
Since $H_2(X)=\mathbb Z$, generated by the whole surface, I described a triangulation $T$ of the double torus, which is a cycle, since it has empty boundary (assuming I checked correctly). Then I considered the cochain $f_a$ defined by $a \to 1$; $\quad b, c, d \to 0$; similarly for $f_b, f_c, f_d, f_T$. Computing the table for the cup product yields the following:
$$\begin{array}{c|c|c|c|c|} 
 \cup & f_a & f_b & f_c & f_d & f_T \\ \hline
f_a & 0 & -f_T & -f_T & -f_T & 0 \\ \hline
f_b & f_T & 0 & -f_T & -f_T & 0 \\ \hline
f_c & f_T & f_T & 0 & -f_T & 0 \\ \hline
f_d & f_T & f_T & f_T & 0 & 0 \\ \hline
f_T & 0 & 0 & 0 & 0 & 0 \\ \hline
\end{array}$$
Is this correct?
 A: That looks right to me (up to a sign convention). The way I checked it is to use Poincare duality, which relates cup product to signed intersection number: look at the vertex $v \in X$ that is the result of gluing the eight corners of the octagon, then look at the four oriented loops $L_a,L_b,L_c,L_d \subset X$ that pass through $v$ and that come from gluing each of the four side pairs $a,b,c,d$, and then check the right-hand rule. It looks to me like you are using a sign convention where the "right hand rule" yields coefficient $-1$ and the "left hand rule" yields the coefficient $+1$. So, for example, the pair $L_a$,$L_b$ passes through $v$ obeying the right hand rule, which gives coefficient $-1$, hence $[f_a] \cup [f_b] = [-f_T]$ (I'm using square braces to denote cohomology classes of cocycles).
Regarding what may or may not look weird, I should add that different gluing patterns are going to give different matrices for the cup product. Your gluing pattern gives this matrix
$$\begin{pmatrix}
0 & -1 & -1 & -1 \\
1 & 0 & -1 & -1 \\
1 & 1 & 0 & -1 \\
1 & 1 & 1 & 0
\end{pmatrix}
$$
whereas the diagram on the right of the post in your question will give this  matrix (which perhaps looks less weird):
$$\begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}
$$
