The exactness of the associated chain complex of a simplicial free abelian group over a finite set and the normalization theorem

Question

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 $n=0,N-1$.

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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?

Answer 1

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.

Answer 2

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.