Consider the above representation of the torus $X$. I need to show that if $\phi\in C^1(X;\mathbb Z)$ is the cochain that takes the value $1$ on the red lines with the orientation given by the vertices (written $v_i$ in the following), then it is a cocycle that is not a coboundary.
In particular, $\phi$ is defined to take the following values, where $[v_i,v_j]$ means the edge bounded by those vertices:
$\phi [v_0,v_1]=1$
$\phi [v_0,v_2]=0$
$\phi [v_1,v_2]=-1$
$\phi [v_1,v_3]=0$
$\phi [v_2,v_3]=1$
We have that $\delta \phi [v_0,v_1,v_2]=\phi[v_1,v_2]-\phi[v_0,v_2]+\phi[v_0,v_1] =-1-0+1=0$, and similarly $\delta \phi [v_1,v_2,v_3]=0$. Hence, $\phi$ is a cocycle.
Now assume for contradiction that $\delta\alpha=\phi$ for some cochain $\alpha\in C^0(X,\mathbb Z)$. Then $\delta\alpha$ takes on the values listed above. Since $\delta\alpha[v_i,v_j]=\alpha(v_j)-\alpha(v_i)$, then we get the relations $\alpha(v_0)=\alpha(v_2)=\alpha(v_1)-1=\alpha(v_3)-1$, which is a perfectly valid cochain. Where is the contradiction?
Best Answer
Note that $v_0,v_1,v_2,$ and $v_3$ are actually all the same vertex, since they are identified when opposite sides of your square are identified. So you cannot have a $0$-cochain that takes different values on them.