How Does Vertex Labeling Affect Chains? I am working through Hatcher on my own and have gotten near the end of Chapter 2.  One thing that I remain confused about are the powers of -1 that are inserted in the formulas for boundaries and other chains which are defined as formal sums.  (See Hatcher p.104 for the first mention of such signs.)
Obviously, these signs are critical in much of what follows since they determine when chains sum to zero.
Here's my problem: The labels/subscripts of the vertices are arbitrary, and thus the signs in the sums are arbitrary.  Consequently, how can we be sure the signs are "right" and, for example, sum to zero for cycles?
Hatcher does not not go into this question.  So, is there some "natural" way of labeling vertices that makes it all work?  Or, is there a set of rules for labeling that always works as long as such rules are followed?
Also, how do we know a "correct" labeling system will always exist as we move to higher dimensions and more complex shapes?
Thanks for any comments or observations.
 A: In short, it doesn’t really matter where the pluses and minuses go. Just as the labels are arbitrary, so is the assignment of orientations—there’s no “right” assignment and it all works out. The orientations need not even be consistent among different-dimension chains within the complex. In fact, it might not even be possible to make such a globally consistent assignment.  
Looking at it from a linear-algebraic point of view, when you assign orientations, you’re choosing a basis for the vector space of chains. The chains that sum to zero are the kernel of the boundary map $\partial_k:C_{k+1}\to C_k$ from $(k+1)$-chains to $k$-chains. This kernel is independent of the choice of bases: changing bases only changes its representation and that of the boundary operator as coefficient matrices.  
Let’s look at what happens with a graph. The boundary operator $\partial:C_1\to C_0$ can be represented as a matrix with rows indexed by node and columns indexed by edge. Relabeling the edges permutes the columns of this matrix, which of course has no effect whatsoever on its kernel and image. Relabeling nodes permutes the rows, which is the same as permuting the coordinates of the tuples that represent chains. So far, so good: the labels are arbitrary and have no effect on which chains sum to zero.  
Now, what happens if you reverse the orientation of edge $j$? This amounts to negating the $j$th column of the matrix $\partial$ or equivalently, of choosing $-\mathbf v_j$ instead of $\mathbf v_j$ as the $j$th basis vector of $C_1$. This simply negates the $j$th coordinate of the tuples that represent $1$-chains, i.e., it changes their representation as coordinate tuples, but not the vectors that those tuples represent.  
One other thing to look at is the effect of changing the orientation of an elementary $k$-chain on $\partial_k$ when $k\gt0$. In matrix terms, doing so negates a row of the matrix of $\partial_k$. This leaves the row space of this matrix unchanged, and since its kernel is the orthogonal complement of its row space, that, too, is unaffected.  
To take a relatively simple example, consider the directed graph with edges $1\to2$, $2\to3$, $1\to3$, $2\to4$ and $3\to4$. The corresponding boundary operator matrix is $$\partial = \begin{bmatrix} -1&0&-1&0&0 \\ 1&-1&0&-1&0 \\ 0&1&1&0&-1 \\ 0&0&0&1&1 \end{bmatrix}.$$ Its kernel is easily found via standard techniques to be spanned by $(-1,-1,1,0,0)^T$ and $(0,1,0,-1,1)^T$ (these are the two smaller loops in the graph), so every linear combination of these two chains sums to zero. For instance, the larger outer loop is the sum of these two loops. Now reverse the orientation of the second edge and recompute: the kernel is now spanned by $(-1,1,1,0,0)^T$ and $(0,-1,0,-1,1)^T$. Comparing these two bases, we find that the only difference is that the second component of the coordinate vectors has been negated, exactly as expected.
