Orientation and simplicial homology I'm reading Chapter 2 of Hatcher's Algebraic Topology, and I just can't figure out the computations of the boundary homomorphism for the examples provided. To provide some context, reproduced the figure for the torus from the book below:

As I understand it, to compute $\partial U$ we follow the faces (which are edges) counter-clockwise, negating an edge if the oriented arrow is "facing us." But starting in the top right corner and working around U results in $\partial U = (-1)^0 (-b) + (-1)^1 (-a) + (-1)^2 c = a - b + c$, which contrasts with the book's result of $\partial U = a + b - c$. I seem to be making some critically flawed assumptions. What am I not understanding?
 A: I don't understand your description of the boundary map, or your explicit calculation.  Here is how the calculation goes:
Imagine that you are standing in $U$, making a counterclockwise pivot, and looking at the boundary (in the naive sense) as you do so.  Let's begin facing the top right corner.  As we turn, our field of vision sweeps out $b$, but in the opposite direction to its arrow (we sweep out $b$ from right to left, while the arrow points from left to right), then $a$, again in the opposite direction to its arrow, and finally $c$, in the same direction as its arrow.
So $\partial U = -b - a + c$.  (If the text instead writes it as $a + b - c$, it must use the opposite orientation on $U$, i.e. a clockwise, rather than counterclockwise, orientation.)
The same procedure applied in $L$, starting by facing the lower right corner, yields $\partial L = a - c + b$, which is $- \partial U$, as one would expect.  (When you glue the $U$ and $L$ into a torus, the boundaries get glued to together, and so "cancel" one another.)
