# Boundary of a simplex

I've been wondering why the boundary of a simplex $\sigma : C_q (X) \rightarrow C_{q-1}(X)$ is defined to be $\partial \sigma = \sum (-1)^i \sigma \circ f_{i,q}$ with alternating sign.

Why can it not be the sum over all faces without alternating sign? Thanks for your help!

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You could of course get rid of the $(-1)^i$, but then you'd have to change the parametrizations $f_{i,q}$ to get the right boundary orientation. If you know some basic manifold theory, consider the simplex to be a manifold (ignore the corners for now) and compute the orientation-form for the boundary simplices. Then, using the parametrization of the boundary simplicies given by $f_{i,q}$, determine how the manifold-theoretic boundary orientation compares to the $f_{i,q}$ orientation. You'll get your formula. – Ryan Budney Jul 31 '11 at 17:12
What do you mean by the parametrizations of $f_{i,q}$? I know nothing about manifolds yet unfortunately. – Rudy the Reindeer Aug 4 '11 at 7:59
All the definitions in simplicial homology were inspired by corresponding notions among manifolds. In particular, if you have a manifold equipped with an orientation, there is a canonical orientation associated to its boundary. Orientations can also be assigned to objects via explicit parametrization by other oriented objects. The formula you're interested in is a combination of these two things. I suggest reading a book on manifolds that covers the general Stokes theorem (using differential forms). Once you understand the proof of the theorem, you'll be ready to reinterpret that formula. – Ryan Budney Aug 4 '11 at 19:02

Because we want to have $\partial\circ\partial=0$ to built homology.
OK, and what about having coefficients in $\mathbb{Z}$? That work, too, right? – Rudy the Reindeer Jul 31 '11 at 11:49
Cources start with $\mathbb{Z}$ but later they extend to any abelian group $G$ or generalize more. You say then that it is "homology with coeffitients" if it is not necessarily $\mathbb{Z}$. – old Jul 31 '11 at 11:58