In computing $\partial: C_2(T) \to C_1(T)$ of the torus, why is $\partial(U)=a+b-c$? Why is the cell structure of the torus oriented this way?

Question

In Example 2.3 from Hatcher, page 106, we are computing the homology groups of the torus. The $2$-chains is the free abelian group $\mathbb Z\{U,L\}$ and the $1$-chains is the free abelian group $\mathbb Z\{a,b,c\}$. Then it is computed that $\partial(U)=\partial(L)=a+b-c$.

I can see how it was computed using the picture, but why is the $c$ edge oriented in that way? Why not have it go the other direction?

I have asked a similar question here, but have not gotten any useful answers.

Answer 1

This is basically just a definition for the $c$-edge. The boundaries of $U$ and $L$ need to be closed. This requires for $L$ (starting from the lower left corner) following $b$ in positive direction, $a$ in positive direction and $c$ in negative direction. Hatcher could have easily used a different orientation for $c$.

Answer 2

$c$ cannot go in the other direction in the picture because the boundary of the $2$-simplex must have the induced orientation.

When drawing a $2$-simplex, just label the vertices as, say, $0,1,2,$ and the edges must be oriented accordingly: from $0$ to $1,$ from $1$ to $2,$ and from $0$ to $2.$ Thus in the draw you cannot have a $2$-simplex with all its edges oriented forming a cycle.

You can, of course, label the vertices as you want. In the picture, the vertices of $L$ as "numbered" as follows: the vertex $0$ is the left-bottom vertex, $1$ is the right-bottom and $2$ the right-upper vertex. Then $\partial(L)=[1,2]-[0,2]+[0,1]=a-c+b.$

If you change $c$ to go in the other direction, then you must change both $b'$s or both $a$'s too in order to respect this rule.