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Update: Now I know why my method fails. But I still wanna know how to work out the original question, that is to show the exactness of the chain complex $C_*(X)$ except for two positions $n=0,N-1$.


Here is a little problem about the normalization theorem in the theory of simplicial objects.

Given a finite set $X$ with $|X|=N>2$, we can construct a simplicial free abelian group $C_*(X)$ defined as follows: for each $n\geq0$, $C_n$ is defined to be the free abelian group generated by the set of all $(n+1)$-tuples $(x_0,x_1,\dots,x_n)$ of distinct elements in $X$; and set $C_n(X)=0$ for $n\geq N$. The differential map is defined as $d(x_0,x_1,\dots,x_n)=\sum_{i=0}^n(-1)^i(x_0,\dots,\hat{x_i},\dots,x_n)$. An exercise in Weibel's book (see Weibel The K-Book: An introduction to algebraic K-theory, VI.5, Ex.5.1., p.550 or p.35) asserts that $H_n(C_*)=0$ for $n\neq 0,N-1$.

My strategy goes as follows: Consider the unnormalized complex $U_*$ with $U_n(X)$ defined to be the free abelian group generated by the unnormalized $(n+1)$-tuple $(x_0,x_1,\dots,x_n)$ of $n$ elements in $X$ which are allowed to have repetitions whose differentials are defined as above. Then an easy computation shows that the chain complex $U_*$ is acyclic for $n\geq1$(See for example Mac Lane's book Homology, Ex.1 in the end of Section 7, Chapter VIII, p. 238). And the normalization theorem(MacLane Homology, VIII.6, p.236 or the Theorem 3.3 in Normalization or thisQuestion) gives that the two chain complexes $C_*, U_*$ are quasi-isomorphic. So the acyclicity of $U_*$ implies the exactness of $C_*$. But $C_{N}=0, C_{N-1}$ is a free abelian group of rank $N!$ which injects into $C_{N-2}$ which also has rank $N!$. This forces the $(N-2)$-th differential $d_{N-2}:C_{N-2}\to C_{N-3}$ to be the zero map, which is impossible. I'm wondering which step goes wrong?

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  • $\begingroup$ Commenting since I was confused a bit by your notation: the groups $C_*(X)$ form a chain complex, but not a simplicial abelian group. (I know this because any simplicial abelian group that vanishes in high degree is identically zero by the Dold-Kan correspondence.) $\endgroup$ Commented Feb 17, 2022 at 17:35
  • $\begingroup$ And probably your links mention this (or my memory is faulty)--this complex is called the "complex of injective words" and proving that is is highly connected is maybe not so direct. Here is a paper by an author I respect arxiv.org/pdf/1608.04496.pdf with a "simple" proof that looks pretty involved to me. $\endgroup$ Commented Feb 17, 2022 at 17:37
  • $\begingroup$ @JohnWiltshire-Gordon : yes I made a mistake. It's semi-simplicial. Thanks for pointing out. And thanks for the reference where the paper perfectly answers my question! $\endgroup$
    – XYC
    Commented Feb 18, 2022 at 7:45

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Leaving an answer so that this question can be marked resolved.

Your chain complex is called the "complex of injective words" and proving that it is highly connected is maybe not so direct. Here is a paper by an author I respect https://arxiv.org/pdf/1608.04496.pdf with a "simple" proof that looks pretty involved to me.

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The point is that a list $(x_0,\dotsc,x_n)$ is only degenerate in $U_*(X)$ if it has two adjacent entries the same. Thus, when you take the quotient by the degenerate subcomplex, lists like $(x,y,x)$ survive, so you do not get the complex $C_*(X)$ that you first thought of.

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  • $\begingroup$ Thx for clarifying. Now I see the difference. But since now the normalization theorem doesn't work for this problem (in a straight way), how can I fix the strategy to show the exactness of $C_*(X)$ except two mentioned positions? $\endgroup$
    – XYC
    Commented Feb 17, 2022 at 4:35

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