# 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?

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Did you double-check that you correctly copied the labels from the book? It looks to me like $\partial U = c - b - a$ and $\partial L = a - c + b$, so maybe you mixed up $U$ and $L$. Other possibility is orientation - is Hatcher going clockwise? – Neal Jun 8 '12 at 1:49
I edited to include the picture in the post. I hope this is okay. – Grumpy Parsnip Jun 8 '12 at 1:50
I've included the actual picture from the book (I opened the pdf file of the book and took a screenshot of the page). – Zev Chonoles Jun 8 '12 at 1:56
@Neal, I just checked again and I copied the labels correctly. From a figure later on in the book, it shows computations of boundaries with a counterclockwise motion. That, along with the following comment that "the orientations of the two hidden faces are also counterclockwise when viewed from the outside the 3-simplex," seems to me that I should also be going counterclockwise. But it's clear I'm definitely making a mistake somewhere, as he also notes that $\partial U = \partial L$, which only confuses me more. – bananas Jun 8 '12 at 1:58
@ZevChonoles and Jim Conant, thanks for the touch ups! – bananas Jun 8 '12 at 1:59

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.)
Thanks for the explanation! I think my error was in trying to use $\partial_n(\sigma_\alpha) = \sum\limits_i (-1)^i \sigma_\alpha | \left[v_0,\cdots,\hat{v_i},\cdots,v_n\right]$ in some horrible combination with the orientations given by the figure. – bananas Jun 8 '12 at 2:24