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Let $A_\bullet$ be a (non-augmented) simplicial object in an abelian category, with face maps $d_i : A_n \to A_{n-1}$ and degeneracy maps $s_i : A_n \to A_{n+1}$, $0 \le i \le n$, for each $n \ge 0$. Let $C A_\bullet$ be the unnormalised chain complex associated with $A_\bullet$: so $C A_n = A_n$ (as objects) and the boundary operator $\partial : C A_n \to C A_{n-1}$ is defined by $\partial = \sum_{i=0}^{n} (-1)^i d_i$ Let $N A_\bullet$ be the normalised Moore complex, where $N A_n = \bigcap_{i=0}^{n-1} \ker (d_i : A_n \to A_{n-1})$ and the boundary operator $\partial : N A_n \to N A_{n-1}$ is the restriction of the face map $(-1)^n d_n : A_n \to A_{n-1}$. This is evidently a subcomplex of $C A_\bullet$, and there is a chain map $p : C A_\bullet \to N A_\bullet$ defined in degree $n$ by $p_n = q_{n-1} \circ \cdots \circ q_0$ where $q_i = \textrm{id} - s_i \circ d_i$. This is a retraction of the inclusion $N A_\bullet \hookrightarrow C A_\bullet$, and the Dold–Kan theorem tells us there is a short exact sequence of chain complexes $0 \to D A_\bullet \to C A_\bullet \overset{p}{\to} N A_\bullet \to 0$ where $D A_\bullet \to C A_\bullet$ is the inclusion of the subcomplex of degenerate simplices, defined by $D A_n = \sum_{i=0}^{n-1} \textrm{im}(s_i : A_{n-1} \to A_n)$ One then shows the induced maps $H_* (C A_\bullet) \to H_* (N A_\bullet)$ are isomorphisms by constructing appropriate chain homotopies. The long exact sequence of homology $\cdots \to H_{n+1} (D A_\bullet) \to H_n (C A_\bullet) \to H_n (N A_\bullet) \to H_n (D A_\bullet) \to \cdots$ then implies that $H_* (D A_\bullet) = 0$, as one expects.

Question. Can we show that $H_* (D A_\bullet) = 0$ more directly, by e.g. showing that $\textrm{id}_\bullet : D A_\bullet \to D A_\bullet$ is null-homotopic? I expect this should be true, since the corresponding simplicial subobject of $A_\bullet$ is contractible (when augmented with $0$). Goerss and Jardine give an explicit formula for the chain homotopy $\textrm{id}_A \Rightarrow p$ (see Ch. III, Thm 2.4), but an explicit formula for the chain homotopy $\textrm{id}_{D A} \Rightarrow 0$ has eluded me.

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Apparently, I already had the answer I wanted but did not realise it: the homotopy $\textrm{id}_{A} \Rightarrow p$ restricts to a homotopy $\textrm{id}_{D A} \Rightarrow 0$.

More generally, suppose we have a direct sum decomposition of chain complexes $C_\bullet = D_\bullet \oplus N_\bullet$ with projections $p_\bullet : C_\bullet \to N_\bullet$ and $r : C_\bullet \to D_\bullet$. If we have a chain homotopy $h : \textrm{id}_{C} \Rightarrow p$, so that $\partial_{n+1} \circ h_n + h_{n-1} \circ \partial_n = \textrm{id}_{C_n} - p_n = r_n$ then $r_{n+1} \circ h_n : C_n \to D_{n+1}$ satisfies $\partial_{n+1} \circ r_{n+1} \circ h_n + r_n \circ h_{n-1} \circ \partial_n = r_n \circ r_n = r_n$ And so, by restricting to $D_n$, we obtain the chain homotopy $\overline{h} : \textrm{id}_D \Rightarrow 0$ as desired.

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In Weibel's book on homological algebra, in the proof of theorem 8.3.8, there is a direct proof that $DA$ is acyclic. It is very natural: one filters the complex with the $p$th layer of $DA_q$ being $F_pD A_q = \sum_{i=0}^{p} \textrm{im}(s_i : A_{n-1} \to A_n)$ and looks at the corresponding spectral sequence. He then shows that the $E_0$ terms of the spectral sequence are acyclic by exhibiting an actual contraction.

This does not (immediately!) give a contraction of $DA$, though.

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    But for this particular purpose you need a very simple minicase of the theory of spectral sequences: that if a the associated graded complex of a filtered complex is exact, then the complex itself is exact.2012-02-11