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We know that the sphere $S^n$ has $n$-th singular homology $H_n(S^n)= \mathbb{Z}$. A generator is given by a fundamental class, which is nothing else than the sum of the simplices in some triangulation. Thus my question is: Is there any intuition behind the fact that such a sum of simplices generates the whole homology? Is there any picture one can have in mind to see every element of the singular homology is a power of this fundamental class?

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The idea is simply this. Suppose you have (a representative of) a nonzero top homology class. Then it has $k\not= 0$ copies of some n-simplex. Since this is a manifold, each boundary face is attached to the boundary face of exactly one other n-simplex, so in order for our chain to be a cycle, we must have $k$ copies of those simplices too. These, in turn, force their neighbors to be in our chain with multiplicity $k$, and so everything propagates around the entire (connected component of) the manifold until we see that this is just $k$ copies of the fundamental class...

Or we get a contradiction. Precisely in the case that we're working with a nonorientable manifold, we'll be able to come back around to our original simplex and we'll want to say that it has multiplicity $-k$, which is impossible unless $k=0$. This illustrates that the top integral homology of a nonorientable manifold is 0, and also suggests why you get a fundamental class for any manifold when you use $\mathbb{Z}/2$ coefficients.

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This is more of a proof than a piece of intuition :) – Mariano Suárez-Alvarez Dec 4 '11 at 1:27
Haha. Well, math makes the most sense when the proof gives intuition! (And, I suppose one could argue, if your proof doesn't give intuition then you've got the wrong proof!) – Aaron Mazel-Gee Dec 4 '11 at 2:12

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