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Let $X$ denote a set. Let $C_{n}(X)$ denote the free abelian group generated by $(n+1)$-tuples of elements of $X$. Define $$\partial_n (x_0, x_1, \ldots, x_n) = \sum_{k=0}^n (-1)^k (x_0,x_1, \ldots, \widehat{x_k}, \ldots, x_n).$$ It is not difficult to show that $\partial_{n-1}\circ \partial_n =0$. I am reading an author who claims that it is clear that this complex is acyclic. While I believe that it is true that it is acyclic, I don't see that it is clear.

It seems to me that one needs to explicitly find a basis for the kernel and determine elements in the image of the previous differential. I can do so for a small set $X$ and for small values of $n$, but I have yet to see a general pattern in either the kernel or the image.

If it is true in general, then there must be a nice reason for it, but I don't get it yet.

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I fixed it, thanks. –  Scott Carter Jan 26 '12 at 10:58

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up vote 2 down vote accepted

The complex in question is the cellular chain complex of $|X|$-dimensional simplex — no wonder it's acyclic.

Algebraically speaking, one can construct (chain) contraction $K\colon C_n\to C_{n+1}$ (let's fix some element $x_0\in X$) "add $x_0$, if it's not in the tuple already, $0$ otherwise" and it's not hard to check that $\partial K\pm K\partial=\operatorname{Id}$ (which implies aciclicity: any cycle $x$ is the boundary of $Kx$).

To what extent that's "clear" depends on your familiarity with algebraic topology / homological algebra, of course (I'd prefer "well-known").

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Thanks Grigory! The map $K$, then is really $K_x$ for some fixed $x$, right? Because I don't ordinarily compute these things (and I need to) I want to check the details before the confirmation of correct. –  Scott Carter Jan 26 '12 at 11:44
@Scott $K$ depends on choice of an element $x_0\in K$, if that's what you mean (general statement is that any cone is contractible and a simplex is a cone over any of it's vertices -- so one need to fix some choice). –  Grigory M Jan 26 '12 at 11:48

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