Introduction
A simplicial set $X$ gives rise to several chain complexes: the unnormalized chain complex $\mathrm{C}_\bullet(X)$, which uses all simplices; the normalized chain complex $N_\bullet(X)$, obtained by quotienting out the degenerate simplices; and the Moore normalized chain complex $N\mathrm{C}_\bullet(X)$, the subcomplex cut out by the face maps $d_0, \ldots, d_{n-1}$. The aim of this note is to show, by an elementary and explicit argument, that these complexes carry the same homological information.
Simplicial Objects
We recall the definition of the simplex category and simplicial objects.
The simplex category $\Delta$ is the category whose objects are the finite ordinals
$$[n] := \{0 < 1 < \cdots < n\}, \quad n \geq 0,$$and whose morphisms are order-preserving (non-decreasing) maps $[m] \to [n]$. Among these, the coface maps $d^i \colon [n-1] \to [n]$ ($0 \leq i \leq n$) are the unique injections omitting $i$, and the codegeneracy maps $s^i \colon [n+1] \to [n]$ ($0 \leq i \leq n$) are the unique surjections sending both $i$ and $i+1$ to $i$.
Let $\mathbf{C}$ be a category. A simplicial object in $\mathbf{C}$ is a functor $X \colon \Delta^{\mathrm{op}} \to \mathbf{C}$. We write $X_n := X([n])$ for the object of $n$-simplices. The induced morphisms
$$d_i := X(d^i) \colon X_n \to X_{n-1}, \quad s_i := X(s^i) \colon X_n \to X_{n+1}$$are called face maps and degeneracy maps, respectively. They satisfy the simplicial identities:
$$ \begin{aligned} & d_k d_j = d_{j-1} d_k \quad (k\leqslant j-1),\\ & s_i s_j = s_{j+1} s_i \quad (i\leqslant j),\\ & d_k s_i = \begin{cases} s_{i-1} d_k, & k\leqslant i-1,\\ \mathrm{id}, & k=i \text{ or } k=i+1,\\ s_i d_{k-1}, & k \geqslant i+2. \end{cases} \end{aligned} $$A morphism of simplicial objects is a natural transformation $f \colon X \to Y$. The resulting category of simplicial objects in $\mathbf{C}$ is denoted $\mathbf{sC}$.
A simplicial object in the category of sets is called a simplicial set, and a simplicial object in the category of abelian groups is called a simplicial abelian group. They are denoted by $\mathbf{sSet}$ and $\mathbf{sAb}$, respectively.
Homology
Let $X$ be a simplicial set. The unnormalized chain complex $\mathrm{C}_\bullet(X)$ is defined by
$$\mathrm{C}_n(X) := \mathbb{Z}[X_n],$$the free abelian group with basis $X_n$, together with boundary maps $\partial_n \colon \mathrm{C}_n(X) \to \mathrm{C}_{n-1}(X)$ given on basis elements by
$$\partial_n(\sigma) := \sum_{i=0}^{n} (-1)^i \, d_i(\sigma), \quad \sigma \in X_n.$$The condition $\partial_{n-1} \circ \partial_n = 0$ follows from the simplicial identities.
We have the composition of functors
$$\Delta^{\mathrm{op}} \xrightarrow{X} \mathbf{Set} \xrightarrow{\mathbb{Z}[-]} \mathbf{Ab},$$which gives a simplicial abelian group. The unnormalized chain complex $\mathrm{C}_\bullet(X)$ is just the associated chain complex of this simplicial abelian group.
The $n$-th simplicial homology group of $X$ is
$$H_n(X) := \frac{\operatorname{Ker} \partial_n}{\operatorname{Im} \partial_{n+1}}.$$Elements of $\ker \partial_n$ are called $n$-cycles and elements of $\operatorname{im} \partial_{n+1}$ are called $n$-boundaries.
We will see later that the homology groups of the unnormalized chain complex $\mathrm{C}_\bullet(X)$ and the normalized chain complex $N_\bullet(X)$ are isomorphic, so we can also compute $H_n(X)$ using $N_\bullet(X)$ without concern about the choice of chain complexes. We follow Goerss–Jardine.
The degenerate subcomplex $D_\bullet(X) \subseteq \mathrm{C}_\bullet(X)$ is defined by
$$ \begin{aligned} D_n(X) &:= \mathbb{Z}\big[\{\, s_j(\sigma) \mid \sigma \in X_{n-1},\; 0 \leq j \leq n-1 \,\}\big] \\ &= \bigcup_{i = 0}^{n-1} \operatorname{Im}(s_i), \end{aligned} $$the subgroup of $\mathrm{C}_n(X)$ generated by all degenerate $n$-simplices. The simplicial identities imply $\partial_n(D_n(X)) \subseteq D_{n-1}(X)$, so $D_\bullet(X)$ is indeed a subcomplex of $\mathrm{C}_\bullet(X)$.
The normalized chain complex $N_\bullet(X)$ is the quotient complex
$$N_n(X) := \mathrm{C}_n(X)\,/\,D_n(X),$$with the boundary map induced by $\partial_n$.
The Moore normalized chain complex is the subcomplex $N\mathrm{C}_\bullet$ of $\mathrm{C}_\bullet$ given by
$$N\mathrm{C}_n = \bigcap_{i = 0}^{n-1} \ker(d_i).$$The boundary map $N\mathrm{C}_{n} \rightarrow N\mathrm{C}_{n-1}$ for each $n$ is given by $(-1)^n d_n$.
Note that by definition of the Moore normalized chain complex, the boundary maps $N\mathrm{C}_{n} \rightarrow N\mathrm{C}_{n-1}$ can also be given by $\partial_n$.
Comparing the Two Complexes
For each $j$ we define the subgroup $N_j\mathrm{C}_n$ of $\mathrm{C}_n$ by
$$N_j \mathrm{C}_n= \begin{cases} \bigcap_{i=0}^{j-1} \ker\left(d_i\right) & \text{for } j \leqslant n-1, \\ N \mathrm{C}_n & \text{for } j \geqslant n, \end{cases}$$and set $N_0\mathrm{C}_n = \mathrm{C}_n$.
As a direct observation, we have $N_{j+1} \mathrm{C}_n \subseteq N_j \mathrm{C}_n$ for all $n$ and $N \mathrm{C}_n = \bigcap_{j = 0}^\infty N_j \mathrm{C}_n$. Hence we have the natural inclusion map $i_j: N_{j+1} \mathrm{C}_n \rightarrow N_{j} \mathrm{C}_n$. There is another group homomorphism $f_j: N_{j} \mathrm{C}_n \rightarrow N_{j+1} \mathrm{C}_n$ given by
$$f_j(x) = \begin{cases} x - s_{j}d_j(x) & \text{for } j \leqslant n-1, \\ x & \text{for } j \geqslant n. \end{cases}$$It is straightforward to check that each $f_j$ indeed takes values in $N_{j+1} \mathrm{C}_n$.
Let $x \in N_{j+1} \mathrm{C}_n$. Then $x \in \bigcap_{i=0}^{j} \ker\left(d_i\right)$ for $j \leqslant n-1$, or $x \in N\mathrm{C}_n$ for $j \geqslant n$. Therefore
$$f_j i_j(x) = f_j(x) = \begin{cases} x - s_{j}d_j(x) = x-0 = x & \text{for } j \leqslant n-1, \\ x & \text{for } j \geqslant n, \end{cases}$$so $f_j i_j(x) = x$ in both cases.
□We can see from the above proposition that $N_{j+1} \mathrm{C}_n$ is a direct summand of $N_{j} \mathrm{C}_n$ and $f_j$ is the projection onto $N_{j+1} \mathrm{C}_n$.
For each fixed $j$, we define a group homomorphism $t_j: N_j\mathrm{C}_n \rightarrow N_j\mathrm{C}_{n+1}$ by
$$t_j(x) = \begin{cases} (-1)^j s_j(x) & \text{for } j \leqslant n, \\ 0 & \text{for } j \geqslant n+1. \end{cases}$$The maps $t_j: N_j\mathrm{C}_n \rightarrow N_j\mathrm{C}_{n+1}$ give a chain homotopy between the identity morphism $\operatorname{id}: N_j \mathrm{C}_n \rightarrow N_j \mathrm{C}_n$ and $i_jf_j: N_j \mathrm{C}_n \rightarrow N_j \mathrm{C}_n$. That is,
$$\partial t_j + t_j \partial = \operatorname{id} - i_jf_j.$$We compute $\partial t_j + t_j \partial$ on $N_j \mathrm{C}_n$. For $j \leqslant n-1$, using the simplicial identities and the assumption that $d_k(x) = 0$ for all $k \in \{0,1, \ldots , j-1\}$, we have
$$ \begin{aligned} (\partial t_j + t_j \partial) (x) & = \sum_{k = 0}^{n+1} (-1)^k d_k (-1)^j s_j(x) + (-1)^j s_j \sum_{k = 0}^{n}(-1)^k d_k (x) \\ & = \sum_{k = 0}^{n+1}(-1)^{j+k} d_k s_j(x) + \sum_{k = 0}^{n}(-1)^{j+k} s_j d_k (x) \\ & = \sum_{k = 0}^{j-1} (-1)^{j+k} s_{j-1} d_k (x) + \sum_{k = j}^{j+1} (-1)^{j+k} x \\ & \qquad + \sum_{k = j+2}^{n+1} (-1)^{j+k} s_jd_{k-1}(x) + \sum_{k = 0}^{n}(-1)^{j+k} s_j d_k (x) \\ & = 0 + 0 + \sum_{k = j+1}^{n} (-1)^{j+k+1} s_jd_k(x) + \sum_{k = j}^{n}(-1)^{j+k}s_j d_k (x) \\ & = s_jd_j (x) + \sum_{k = j+1}^{n} \bigl((-1)^{j+k+1} + (-1)^{j+k}\bigr) s_jd_k(x) \\ & = s_jd_j (x) \\ & = x - f_j(x) = (\operatorname{id} - i_jf_j)(x). \end{aligned} $$Here the first sum vanishes because $d_k(x) = 0$ for $k \leqslant j-1$, the two terms with $k = j, j+1$ cancel, and in the last sum of the third equality the terms with $k \leqslant j-1$ vanish for the same reason.
For $j \geqslant n$, we have $\operatorname{id} - i_jf_j = 0$ on $N_j\mathrm{C}_n$. If $j \geqslant n+1$ then $t_j = 0$ in degrees $n$ and $n-1$, so $\partial t_j + t_j\partial = 0$. If $j = n$, then $t_j\partial(x) = 0$ since $\partial(x) \in N_j\mathrm{C}_{n-1}$ and $j \geqslant (n-1)+1$, while $\partial t_j(x) = (-1)^n \sum_{k=0}^{n+1} (-1)^k d_k s_n(x) = 0$, because the terms with $k \leqslant n-1$ equal $s_{n-1}d_k(x) = 0$ and the terms with $k = n, n+1$ cancel. Hence $\partial t_j + t_j \partial = \operatorname{id} - i_jf_j$ in all degrees.
□Therefore we have a series of homotopy equivalences of chain complexes $\{N_j\mathrm{C}_\bullet\}_{j = 0}^{n-1}$:
$$ \begin{aligned} \mathrm{C}_n = N_0\mathrm{C}_n & \;\underset{i_0}{\overset{f_0}{\rightleftarrows}}\; N_1\mathrm{C}_n \;\underset{i_1}{\overset{f_1}{\rightleftarrows}}\; N_2\mathrm{C}_n \;\underset{i_2}{\overset{f_2}{\rightleftarrows}}\; \cdots \;\underset{i_{n-2}}{\overset{f_{n-2}}{\rightleftarrows}}\; N_{n-1}\mathrm{C}_n = N\mathrm{C}_n \end{aligned} $$Furthermore, for each $j$, the composition $f_{j-1} \cdots f_0$ is a projection from $\mathrm{C}_n$ onto $N_j\mathrm{C}_n$. We write
$$\mathrm{C}_n = N_j\mathrm{C}_n \oplus D_j\mathrm{C}_n, \quad j = 0,1,\ldots, n-1,$$where $D_j\mathrm{C}_n = \ker(f_{j-1}\cdots f_0)$ and $N_j\mathrm{C}_n \cap D_j\mathrm{C}_n = \{0\}$. In particular, $f_{n-2} \cdots f_1 f_0$ is a projection from $\mathrm{C}_n$ onto $N\mathrm{C}_n$, and the complementary subgroup is just $\ker(f_{n-2} \cdots f_1 f_0)$.
We first prove $D_n \subseteq \ker(f_{n-2} \cdots f_1 f_0)$ by repeatedly using the simplicial identities. If $s_j(y) \in D_n$, then
$$f_j(s_j(y)) = s_j(y) - s_jd_j(s_j(y)) = s_j(y) - s_j(y) = 0,$$and for $j \geqslant i + 1$,
$$ \begin{aligned} f_i(s_j(y)) &= s_j(y) - s_id_i(s_j(y)) = s_j(y) - s_is_{j-1}d_i(y) \\ &= s_j(y) - s_j(s_id_i(y)) = s_j(y - s_id_i(y)) = s_j(f_i(y)), \end{aligned} $$meaning that we can swap $s_j$ and $f_i$ in this setting. Therefore, for $0 \leqslant j \leqslant n-2$,
$$f_{n-2} \cdots f_1 f_0 (s_j(y)) = f_{n-2} \cdots f_j\, s_j(f_{j-1} \cdots f_0(y))=0.$$Now we prove $\ker(f_{n-2} \cdots f_1 f_0) \subseteq D_n$. If $x \in \ker(f_{n-2} \cdots f_1 f_0)$, then
$$ \begin{aligned} 0 &= f_{n-2} \cdots f_0 (x) \\ &= (\operatorname{id} - s_{n-2}d_{n-2}) \cdots (\operatorname{id} - s_{1}d_{1})(\operatorname{id} - s_{0}d_{0})(x) \\ &= x - s_{n-2}(\text{something}). \end{aligned} $$Hence $x = s_{n-2}(\text{something}) \in D_n$.
□Since $f_{n-2}\cdots f_0 : \mathrm{C}_n \rightarrow N\mathrm{C}_n$ is a surjective group homomorphism, we have
$$N_n = \mathrm{C}_n / D_n = \mathrm{C}_n / \ker(f_{n-2} \cdots f_0) \cong N\mathrm{C}_n$$by the first isomorphism theorem. Moreover, these isomorphisms are compatible with the boundary maps, i.e. the following diagram commutes:
$$ \begin{array}{ccc} N\mathrm{C}_{n} & \xrightarrow{(-1)^nd_n} & N\mathrm{C}_{n-1} \\ \downarrow \wr & & \downarrow \wr \\ N_{n} & \xrightarrow{\;\partial_n\;} & N_{n-1} \end{array} $$□Summing up, we have the following result:
References
- Goerss, P. G. and Jardine, J. F. (2009) Simplicial Homotopy Theory. Modern Birkhäuser Classics. Basel: Birkhäuser Verlag. Reprint of the 1999 edition. ISBN 978-3-0346-0188-7. Available at: https://doi.org/10.1007/978-3-0346-0189-4.