# How is the generator of the first homology of the torus non-trivial?

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?

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.