# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

### 2.5.6 The Dold-Kan Correspondence

Let $\operatorname{ Ab }$ denote the category of abelian groups, and $\operatorname{ Ab }_{\Delta } = \operatorname{Fun}( \operatorname{{\bf \Delta }}^{\operatorname{op}}, \operatorname{ Ab })$ the category of simplicial abelian groups. The formation of normalized Moore complexes (Construction 2.5.5.7) determines a functor $\mathrm{N}_{\ast }: \operatorname{ Ab }_{\Delta } \rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}})$. Our goal in this section is to prove the following fundamental result, which was discovered independently by Dold () and Kan ():

Theorem 2.5.6.1 (The Dold-Kan Correspondence). The normalized Moore complex functor determines an equivalence of categories $\mathrm{N}_{\ast }: \operatorname{ Ab }_{\Delta } \rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}})_{\geq 0}$.

Remark 2.5.6.2. Theorem 2.5.6.1 admits many generalizations. For example, if $\operatorname{\mathcal{A}}$ is an abelian category (Definition ), then a variant of Construction 2.5.5.9 supplies an equivalence of categories

$\mathrm{N}_{\ast }: \{ \text{Simplicial objects of \operatorname{\mathcal{A}}} \} \rightarrow \operatorname{Ch}(\operatorname{\mathcal{A}})_{\geq 0},$

where $\operatorname{Ch}(\operatorname{\mathcal{A}})_{\geq 0}$ denotes the category of (nonnegatively graded) chain complexes with values in $\operatorname{\mathcal{A}}$ (see Theorem ). For more general categories $\operatorname{\mathcal{A}}$, one can think of the category of simplicial objects $\operatorname{\mathcal{A}}_{\Delta } = \operatorname{Fun}( \operatorname{{\bf \Delta }}^{\operatorname{op}}, \operatorname{\mathcal{A}})$ as a replacement for the category of chain complexes $\operatorname{Ch}(\operatorname{\mathcal{A}})_{\geq 0}$, which is better behaved in “non-additive” situations.

We begin by constructing a right adjoint to the normalized Moore complex functor.

Construction 2.5.6.3 (The Eilenberg-MacLane Functor). Let $n$ be a nonnegative integer and let $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ denote the normalized chain complex of the standard $n$-simplex (Construction 2.5.5.9). For every chain complex $M_{\ast }$, we let $\mathrm{K}_{n}( M_{\ast } )$ denote the collection of chain maps from $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ into $M_{\ast }$ (which we regard as an abelian group under addition). Note that the construction $[n] \mapsto \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ determines a functor from the simplex category $\operatorname{{\bf \Delta }}$ to the category of chain complexes, so we can regard $[n] \mapsto \mathrm{K}_{n}( M_{\ast } )$ as a functor from $\operatorname{{\bf \Delta }}^{\operatorname{op}}$ to the category of abelian groups. We denote this simplicial abelian group by $\mathrm{K}( M_{\ast } )$, and refer to it as the Eilenberg-MacLane space associated to $M_{\ast }$.

Remark 2.5.6.4. Let $M_{\ast }$ be a chain complex. We will generally not distinguish in notation between the simplicial abelian group $\mathrm{K}( M_{\ast } )$ and its underlying simplicial set. Note that $\mathrm{K}( M_{\ast } )$ is automatically a Kan complex (Proposition 1.1.9.9), which motivates our usage of the term “space”.

Example 2.5.6.5. Let $M_{\ast }$ be a chain complex. Then we have canonical isomorphisms

$\mathrm{K}_{0}( M_{\ast } ) = \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}}), M_{\ast } ) = \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \operatorname{\mathbf{Z}}, M_{\ast }) = \mathrm{Z}_0( M ).$

In other words, we can identify vertices of the simplicial set $\mathrm{K}(M_{\ast })$ with $0$-cycles of the chain complex $M_{\ast }$.

Example 2.5.6.6. Let $M_{\ast }$ be a chain complex, and let $x,y \in M_0$ be a pair of $0$-cycles, which we identify with vertices of the simplicial set $\mathrm{K}(M_{\ast })$. The following conditions are equivalent:

$(a)$

The vertices $x$ and $y$ belong to the same connected component of the simplicial set $\mathrm{K}(M_{\ast } )$ (Definition 1.1.6.8).

$(b)$

There exists an edge $e$ of the simplicial set $\mathrm{K}(M_{\ast })$ connecting $x$ to $y$ (so that $d_1(e) = x$ and $d_0(e) = y$).

$(c)$

The cycles $x$ and $y$ are homologous: that is, there exists an element $u \in M_{1}$ satisfying $\partial (u) = x - y$.

The equivalence of $(a) \Leftrightarrow (b)$ follows from the fact that $\mathrm{K}( M_{\ast } )$ is a Kan complex (see Remark 1.3.6.15), while the equivalence $(b) \Leftrightarrow (c)$ follows immediately from the construction of the simplicial set $\mathrm{K}(M_{\ast } )$. It follows that the set of connected components $\pi _0( \mathrm{K}(M_{\ast } ))$ can be identified with the $0$th homology group $\mathrm{H}_{0}( M )$.

We now describe a particularly important special case of Construction 2.5.6.3.

Construction 2.5.6.7 (Eilenberg-MacLane Spaces). Let $A$ be an abelian group, let $n$ be an integer, and let $A[n]$ denote the chain complex consisting of the single abelian group $A$, concentrated in degree $n$ (Example 2.5.1.2). We will denote the simplicial abelian group $\mathrm{K}( A[n] )$ by $\mathrm{K}(A,n)$ and refer to it as the $n$th Eilenberg-MacLane space of $A$.

Remark 2.5.6.8. The formation of Eilenberg-MacLane spaces $A \mapsto \mathrm{K}(A,n)$ is defined for every integer $n$. However, it is only interesting for $n \geq 0$: if $n$ is negative, then the simplicial abelian group $\mathrm{K}(A,n)$ is trivial (that is, it is isomorphic to $\Delta ^0$ as a simplicial set).

Example 2.5.6.9. Let $A$ be an abelian group. To supply an $n$-simplex of the simplicial set $\mathrm{K}(A,0)$, one must give a chain map $\sigma : \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow A$. By definition, a homomorphism of graded abelian groups from $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ to $A$ is given by a tuple $\{ a_ i \} _{0 \leq i \leq n}$ of elements of $A$, indexed by the set $[n] = \{ 0 < 1 < \cdots < n \}$ of vertices of $\Delta ^ n$. Under this identification, the chain maps can be identified with those tuples $\{ a_ i \} _{0 \leq i \leq n}$ which are constant: that is, which satisfy $a_ i = a_ j$ for all $i,j \in [n]$. It follows that the Eilenberg-MacLane space $\mathrm{K}(A,0)$ can be identified with the constant simplicial abelian group taking the value $A$.

Example 2.5.6.10. Let $A$ be an abelian group. To supply an $n$-simplex of the simplicial set $\mathrm{K}(A,1)$, one must give a chain map $\sigma : \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow A$. By definition, a homomorphism of graded abelian groups from $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ to $A$ is given by a system $\{ a_{i,j} \} _{0 \leq i < j \leq n}$ of elements of $A$, indexed by the set of all nondegenerate edges of $\Delta ^ n$. Under this identification, the chain maps can be identified with those systems $\{ a_{i,j} \} _{0 \leq i < j \leq n}$ satisfying $a_{i,j} + a_{j,k} = a_{i,k}$ for $0 \leq i < j < k \leq n$. It follows that the Eilenberg-MacLane space $\mathrm{K}(A,1)$ can be identified with the the Milnor construction $B_{\bullet } A$ (Example 1.2.4.3).

Notation 2.5.6.11. Let $M_{\ast }$ be a chain complex. Then every $n$-simplex $\sigma$ of the simplicial set $\mathrm{K}( M_{\ast } )$ can be identified with a map of chain complexes $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow M_{\ast }$, which carries the generator of $\mathrm{N}_{n}( \Delta ^ n; \operatorname{\mathbf{Z}})$ to an $n$-chain $\widetilde{v}(\sigma ) \in M_ n$. Moreover:

• Since $\sigma$ is a map of chain complexes, we have

$\partial ( \widetilde{v}(\sigma ) ) = \sum _{i =0}^{n} (-1)^{i} \widetilde{v}( d_ i \sigma ) ).$

In other words, the construction $\sigma \mapsto \widetilde{v}(\sigma )$ determines a chain map from the Moore complex $\mathrm{C}_{\ast }( \mathrm{K}( M_{\ast } ) )$ to the chain complex $M_{\ast }$.

• If $\sigma$ is a degenerate $n$-simplex of $\mathrm{K}( M_{\ast } )$, then the map of chain complexes $\sigma : \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow M_{\ast }$ factors through $\mathrm{N}_{\ast }( \Delta ^{m}; \operatorname{\mathbf{Z}})$ for some $m < n$, and therefore annihilates the generator of $\mathrm{N}_{n}( \Delta ^ n; \operatorname{\mathbf{Z}})$. It follows that $\widetilde{v}$ factors (uniquely) as a composition

$\mathrm{C}_{\ast }( \mathrm{K}( M_{\ast } ) ) \twoheadrightarrow \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast }) ) \xrightarrow {v} M_{\ast }.$

We will refer to the chain map $v: \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast }) ) \rightarrow M_{\ast }$ as the counit map.

Proposition 2.5.6.12. Let $M_{\ast }$ be a chain complex and let $v: \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast }) ) \rightarrow M_{\ast }$ be the counit map of Notation 2.5.6.11. Then, for any simplicial abelian group $A_{\bullet }$, the composite map

$\theta : \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( A_{\bullet }, \mathrm{K}( M_{\ast }) ) \rightarrow \operatorname{Hom}_{\operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), \mathrm{N}_{\ast }( \mathrm{K}( M_{\ast }) ) ) \xrightarrow {v \circ } \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), M_{\ast } )$

is an isomorphism of abelian groups.

Proof. Let us say that a simplicial abelian group $A_{\bullet }$ is free if it can be written as a (possibly infinite) direct sum of simplicial abelian groups of the form $\operatorname{\mathbf{Z}}[ \Delta ^ n ]$. Note that every simplicial abelian group $A_{\bullet }$ admits a surjection $P_{\bullet } \twoheadrightarrow A_{\bullet }$, where $P_{\bullet }$ is free (for example, we can take $P_{\bullet }$ to be the direct sum $\bigoplus _{\sigma } \operatorname{\mathbf{Z}}[ \Delta ^{ \mathrm{dim}(\sigma ) }]$ where $\sigma$ ranges over all the simplices of $A_{\bullet }$). Applying this observation twice, we observe that every simplicial abelian group $A_{\bullet }$ admits a resolution

$Q_{\bullet } \rightarrow P_{\bullet } \twoheadrightarrow A_{\bullet } \rightarrow 0,$

which determines a commutative diagram of exact sequences

$\xymatrix {0 \ar [r] & \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( A_{\bullet }, \mathrm{K}( M_{\ast }) ) \ar [r] \ar [d]^{\theta } & \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( P_{\bullet }, \mathrm{K}( M_{\ast }) ) \ar [d]^{\theta '} \ar [r] & \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( Q_{\bullet }, \mathrm{K}( M_{\ast }) ) \ar [d]^{\theta ''} \\ 0 \ar [r] & \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), M_{\ast } ) \ar [r] & \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(P), M_{\ast } ) \ar [r] & \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(Q), M_{\ast } ). }$

Consequently, to prove that $\theta$ is an isomorphism, it will suffice to show that $\theta '$ and $\theta ''$ are isomorphisms. In other words, we may assume without loss of generality that the simplicial abelian group $A_{\bullet }$ is free. Decomposing $A_{\bullet }$ as a direct sum, we can further reduce to the case $A_{\bullet } = \operatorname{\mathbf{Z}}[ \Delta ^ n ]$, in which case the result follows immediately from the definitions. $\square$

Corollary 2.5.6.13. The normalized Moore complex functor $\mathrm{N}_{\ast }: \operatorname{ Ab }_{\Delta } \rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}})$ admits a right adjoint $\mathrm{K}: \operatorname{Ch}(\operatorname{\mathbf{Z}}) \rightarrow \operatorname{ Ab }_{\Delta }$, given on objects by Construction 2.5.6.3.

Note that we can also regard $M_{\ast } \mapsto \mathrm{K}( M_{\ast } )$ as a functor from chain complexes to simplicial sets (by neglecting the group structure on $\mathrm{K}( M_{\ast } )$. This simplicial set also has a universal property:

Corollary 2.5.6.14. The normalized chain complex functor

$\mathrm{N}_{\ast }( -; \operatorname{\mathbf{Z}}): \operatorname{Set_{\Delta }}\rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}})$

admits a right adjoint, given on objects by the functor $M_{\ast } \mapsto \mathrm{K}( M_{\ast } )$ of Construction 2.5.6.3.

Remark 2.5.6.15. When regarded as a functor from $\operatorname{Ch}(\operatorname{\mathbf{Z}})$ to the category of simplicial sets, the functor $M_{\ast } \mapsto \mathrm{K}( M_{\ast } )$ fits into the paradigm of Variant 1.1.7.6: it is the functor $\operatorname{Sing}^{Q}_{\bullet }$ associated to the cosimplicial chain complex

$Q: \operatorname{{\bf \Delta }}\rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}}) \quad \quad [n] \mapsto \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}).$

To deduce Theorem 2.5.6.1, it is convenient to use a different description of the normalized Moore complex.

Construction 2.5.6.16 (The Normalized Moore Complex: Second Construction). Let $A_{\bullet }$ be a simplicial abelian group. For each $n \geq 0$, we let $\widetilde{ \mathrm{N} }_{n}( A)$ denote the subgroup of $\mathrm{C}_{n}(A) = A_ n$ consisting of those elements $x$ which satisfy $d_{i}(x) = 0$ for $1 \leq i \leq n$. Note that if $x$ satisfies this condition, then we have

$\partial (x) = \sum _{i = 0}^{n} (-1)^{i} d_ i(x) = d_0(x).$

Moreover, the identity $d_{i} d_0(x) = d_0 d_{i+1}(x) = 0$ shows that $\partial (x) = d_0$ belongs to the subgroup $\widetilde{ \mathrm{N} }_{n-1}(A) \subseteq \mathrm{C}_{n-1} = A_{n-1}$. We can therefore regard $\widetilde{N}_{\ast }( A)$ as a subcomplex of the Moore complex $\mathrm{C}_{\ast }(A)$.

In the situation of Construction 2.5.6.16, we will abuse terminology by referring to the chain complex $\widetilde{\mathrm{N}}_{\ast }(A)$ as the normalized Moore complex of $A_{\bullet }$. This abuse is justified by the observation that the chain complexes $\widetilde{\mathrm{N}}_{\ast }(A)$ is canonically isomorphic to the normalized Moore complex $\mathrm{N}_{\ast }(A)$ of Construction 2.5.5.7 (Proposition 2.5.6.19 below). We will deduce this from the following more precise statement:

Lemma 2.5.6.17. Let $A_{\bullet }$ be a simplicial abelian group and let $n$ be a nonnegative integer. Then the map

$f: \bigoplus _{ \alpha : [n] \twoheadrightarrow [m] } \widetilde{\mathrm{N}}_{m}(A) \rightarrow A_{n} \quad \quad \{ x_{\alpha } \} \mapsto \sum \alpha ^{\ast }( x_{\alpha } )$

is an isomorphism of abelian groups. Here the direct sum is indexed by surjective nondecreasing maps $\alpha : [n] \rightarrow [m]$ for $0 \leq m \leq n$, and $\alpha ^{\ast }: A_{m} \rightarrow A_{n}$ denotes the associated group homomorphism.

Proof. We first prove that $f$ is surjective. The proof proceeds by induction on $n$. By virtue of our inductive hypothesis, the image of $f$ contains the subgroups $\widetilde{\mathrm{N}}_{n}(A), \mathrm{D}_{n}(A) \subseteq \mathrm{C}_{n}(A) = A_ n$. It will therefore suffice to show that the composite map

$\widetilde{\mathrm{N}}_{n}(A) \hookrightarrow \mathrm{C}_{n}( A ) \twoheadrightarrow \mathrm{C}_{n}(A) / \mathrm{D}_{n}(A)$

is surjective. Fix an element $\overline{x} \in \mathrm{C}_{n}(A) / \mathrm{D}_{n}(A)$. For each $x \in \mathrm{C}_{n}(A)$ representing $\overline{x}$, let $i_{x}$ be the smallest nonnegative integer such that $d_ j(x)$ vanishes for $i_{x} < j \leq n$. Without loss of generality, we may assume that $x$ is chosen so that $i = i_{x}$ is as small as possible. We wish to prove that $i = 0$ (so that $x$ belongs to $\mathrm{N}_{n}(A)$). Assume otherwise, and set $y = x - (s_{i-1} \circ d_ i)(x)$. Then $y$ is congruent to $x$ modulo $\mathrm{D}_{n}(A)$, and for $i \leq j \leq n$ we have

\begin{eqnarray*} d_ j(y) & = & d_ j(x) - (d_ j \circ s_{i-1} \circ d_ i) )(x) ) \\ & = & d_ j( x) - \begin{cases} d_ i(x) & \text{ if } i = j \\ (s_{i-1} \circ d_{j-1} \circ d_ i )(x) & \text{ if } i < j. \end{cases}\\ & = & d_ j(x) - \begin{cases} d_ i(x) & \text{ if } i = j \\ (s_{i-1} \circ d_{i} \circ d_ j )(x) & \text{ if } i < j. \end{cases}\\ & = & 0. \end{eqnarray*}

It follows that $i_{y} < i = i_{x}$, contradicting our choice of $x$.

We now prove that $f$ is injective. Suppose otherwise, so that there exists a nonzero element

$\{ x_{\alpha } \} \in \bigoplus _{ \alpha : [n] \twoheadrightarrow [m] } \widetilde{\mathrm{N}}_{m}(A)$

which is annihilated by $f$. Then there exists some surjective map $\beta : [n] \twoheadrightarrow [k]$ such that $x_{\beta }$ is nonzero. Assume that $k$ has been chosen as small as possible. Moreover, we may assume that $\beta$ is maximal among nondecreasing maps $[n] \twoheadrightarrow [k]$ such that $x_{\beta } \neq 0$: in other words, that for any other map $\alpha : [n] \twoheadrightarrow [k]$ satisfying $\beta (i) \leq \alpha (i)$ for $0 \leq i \leq n$, we either have $\beta = \alpha$ or $x_{\alpha } = 0$. Let $\gamma : [k] \rightarrow [n]$ be the map given by $\gamma (j) = \min \{ i \in [n]: \beta (i) = j \}$. Then $\gamma$ is a nondecreasing map satisfying $\beta \circ \gamma = \operatorname{id}_{[k]}$ and $\gamma (0) = 0$. We then have

\begin{eqnarray*} \gamma ^{\ast } f( \{ x_{\alpha } \} ) & = & \gamma ^{\ast }( \sum _{ \alpha : [n] \twoheadrightarrow [m] } \alpha ^{\ast }( x_{\alpha } )) \\ & = & \sum _{ \alpha : [n] \twoheadrightarrow [m] } ( \alpha \circ \gamma )^{\ast }( x_{\alpha } ). \end{eqnarray*}

We now inspect the summands appearing on the right hand side:

• Let $\alpha : [n] \twoheadrightarrow [m]$ be a surjective nondecreasing function, and suppose that the composite map $[k] \xrightarrow {\gamma } [n] \xrightarrow {\alpha } [m]$ is not surjective. Then we can choose $0 \leq i \leq m$ such that $i$ does not belong the the image of $\alpha \circ \gamma$. Then the homomorphism $(\alpha \circ \gamma )^{\ast }: A_{m} \rightarrow A_{k}$ factors through the face map $d_ i: A_{m} \rightarrow A_{m-1}$. Note that we must have $i > 0$ (since $\gamma (0) = 0$ and $\alpha (0) = 0$), so that $x_{\alpha }$ is annihilated by $d_ i$ (by virtue of our assumption that $x_{\alpha }$ belongs to the subgroup $\mathrm{N}_{m}(A) \subseteq A_ m$) and therefore also by $(\alpha \circ \gamma )^{\ast }$.

• Let $\alpha : [n] \twoheadrightarrow [m]$ be a surjective nondecreasing function, and suppose that the composite map $[k] \xrightarrow {\gamma } [n] \xrightarrow {\alpha } [m]$ is surjective but not injective. In this case, we must have $m < k$, so that $x_{\alpha }$ vanishes by virtue of the minimality assumption on $k$.

• Let $\alpha : [n] \twoheadrightarrow [m]$ be a surjective map, and suppose that the composite map $[k] \xrightarrow {\gamma } [n] \xrightarrow {\alpha } [m]$ is bijective, so that $m = k$ and $\alpha \circ \gamma$ is the identity on $[k]$. For $0 \leq i \leq n$, we have $(\gamma \circ \beta )(i) \leq i$ (by the definition of $\gamma$), so that

$\beta (i) = ((\alpha \circ \gamma ) \circ \beta )(i) = (\alpha \circ (\gamma \circ \beta ))(i) \leq \alpha (i).$

Invoking our maximality assumption on $\beta$, we conclude that either $\alpha = \beta$ or $x_{\alpha }$ vanishes.

Combining these observations, we obtain an equality

$x_{\beta } = \sum _{ \alpha : [n] \twoheadrightarrow [m] } ( \alpha \circ \gamma )^{\ast }( x_{\alpha } ) = \gamma ^{\ast } f( \{ x_{\alpha } \} ) = 0,$

contradicting our choice of $\beta$. $\square$

Remark 2.5.6.18. Let $f: A_{\bullet } \rightarrow B_{\bullet }$ be a morphism of simplicial abelian groups. By virtue of Lemma 2.5.6.17, the following assertions are equivalent:

• For every integer $n \geq 0$, the map of abelian groups $A_{n} \rightarrow B_ n$ is surjective (respectively split surjective, injective, split injective).

• For every integer $n \geq 0$, the map of abelian groups $\mathrm{N}_{n}(A) \rightarrow \mathrm{N}_{n}(B)$ is surjective (respectively split surjective, injective, split injective).

Proposition 2.5.6.19. Let $A_{\bullet }$ be a simplicial abelian group. Then the composite map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A) \twoheadrightarrow \mathrm{N}_{\ast }(A)$ is an isomorphism of chain complexes. In other words, the Moore complex $\mathrm{C}_{\ast }(A)$ splits as a direct sum of the subcomplex $\widetilde{ \mathrm{N} }_{\ast }(A)$ of Construction 2.5.6.16 and the subcomplex $\mathrm{D}_{\ast }(A)$ of Proposition 2.5.5.6.

Proof. The surjectivity of the composite map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A) \twoheadrightarrow \mathrm{N}_{\ast }(A)$ follows from Lemma 2.5.6.17. Moreover, it follows by induction that the subgroup $\mathrm{D}_{n}(A) \subseteq A_ n$ is generated by the images of the maps

$\widetilde{\mathrm{N}}_{m}(A) \hookrightarrow A_{m} \xrightarrow { \alpha ^{\ast } } A_{n}$

where $\alpha : [n] \twoheadrightarrow [m]$ is a nondecreasing surjection and $m < n$, so that the injectivity of $\rho$ also follows from Lemma 2.5.6.17. $\square$

Warning 2.5.6.20. Let $A_{\bullet }$ be a simplicial abelian group, and let $A_{\bullet }^{\operatorname{op}}$ be the opposite simplicial abelian group (obtained by precomposing the functor $A_{\bullet }: \operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{ Ab }$ with the order-reversal involution $\mathrm{Op}: \operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{{\bf \Delta }}^{\operatorname{op}}$ of Notation 1.3.2.1). Then there is a canonical isomorphism of Moore complexes $\psi : \mathrm{C}_{\ast }( A^{\operatorname{op}} ) \simeq \mathrm{C}_{\ast }(A )$, given by $\psi (x) = (-1)^{n} x$ for $x \in A_ n$. This isomorphism carries the subcomplex $\mathrm{D}_{\ast }( A^{\operatorname{op}} )$ generated by the degenerate simplices of $A_{\bullet }^{\operatorname{op}}$ to the subcomplex $\mathrm{D}_{\ast }( A )$ generated by the degenerate simplices of $A_{\bullet }$, and therefore descends to an isomorphism of normalized Moore complexes $\mathrm{N}_{\ast }( A^{\operatorname{op}} ) \simeq \mathrm{N}_{\ast }( A )$, where we view $\mathrm{N}_{\ast }(A)$ and $\mathrm{N}_{\ast }(A^{\operatorname{op}})$ as quotients of $\mathrm{C}_{\ast }(A)$ and $\mathrm{C}_{\ast }(A^{\operatorname{op}})$ (as in Construction 2.5.5.7). Beware that the isomorphism $\psi$ does not carry the subcomplex $\widetilde{ \mathrm{N} }_{\ast }(A^{\operatorname{op}}) \subseteq \mathrm{C}_{\ast }(A^{\operatorname{op}})$ of Construction 2.5.6.16 to the subcomplex $\widetilde{ \mathrm{N} }_{\ast }(A) \subseteq \mathrm{C}_{\ast }(A)$. Instead, it carries it carries $\widetilde{ \mathrm{N} }_{\ast }( A^{\operatorname{op}} )$ to a different subcomplex of $\mathrm{C}_{\ast }(A)$, given in degree $n$ by those elements $x \in \mathrm{C}_{n}(A) = A_ n$ satisfying $d_ i(x)$ for $0 \leq i < n$, and with differential given by $x \mapsto (-1)^{n} d_ n(x)$. This subcomplex is yet another incarnation of the normalized Moore complex of $A_{\bullet }$, which is canonically isomorphic to $\widetilde{ \mathrm{N} }_{\ast }(A)$ but not identical as a subcomplex of $\mathrm{C}_{\ast }(A)$.

More informally: the definition of the normalized Moore complex $\mathrm{N}_{\ast }(A)$ as a quotient of $\mathrm{C}_{\ast }(A)$ (via Construction 2.5.5.7) is compatible with passage from a simplicial abelian group $A_{\bullet }$ to its opposite $A_{\bullet }^{\operatorname{op}}$, but the realization as a subcomplex of $\mathrm{C}_{\ast }(A)$ (via Construction 2.5.6.16) is not.

Remark 2.5.6.21. Let $A_{\bullet }$ be a simplicial abelian group. Then Warning 2.5.6.20 supplies a canonical isomorphism of normalized Moore complexes $\mathrm{N}_{\ast }(A) \simeq \mathrm{N}_{\ast }( A^{\operatorname{op}} )$. By virtue of Theorem 2.5.6.1, this isomorphism can be lifted uniquely to an isomorphism of simplicial abelian groups $\varphi : A_{\bullet } \simeq A_{\bullet }^{\operatorname{op}}$. The isomorphism $\varphi$ is characterized by the requirement that for every $n$-simplex $x \in A_ n$, we have $\varphi (x) \equiv (-1)^{n} x$ modulo degenerate simplices of $A_{\bullet }$.

We now use Proposition 2.5.6.19 to deduce Proposition 2.5.5.11, which was stated without proof in §2.5.5. The statement can be reformulated as follows:

Proposition 2.5.6.22. Let $A_{\bullet }$ be a simplicial abelian group. Then:

• The quotient map $\mathrm{C}_{\ast }(A) \twoheadrightarrow \mathrm{N}_{\ast }(A)$ induces an isomorphism on homology.

• The inclusion map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A)$ induces an isomorphism on homology.

• The subcomplex $\mathrm{D}_{\ast }(A) \subseteq \mathrm{C}_{\ast }(A)$ of Notation 2.5.5.5 is acyclic: that is, its homology groups are trivial.

Proof. By virtue of Proposition 2.5.6.19, assertions $(a)$, $(b)$, and $(c)$ are equivalent. It will therefore suffice to prove $(b)$. Note that the map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A)$ is the inclusion of a direct summand (Proposition 2.5.6.19) and is therefore automatically injective on homology. To show that it also induces a surjective map, it will suffice to show that every $n$-cycle $x \in \mathrm{C}_{n}(A)$ is homologous to an element of the subgroup $\widetilde{\mathrm{N}}_{n}(A)$. Let $i$ denote the smallest nonnegative integer for which the faces $d_ j(x)$ vanish for $i < j \leq n$; our proof will proceed by induction on $i$. If $i = 0$, then $x$ belongs to $\widetilde{\mathrm{N}}_{n}(A)$, and there is nothing to prove. Otherwise, let $y \in \mathrm{C}_{n}(A)$ denote the boundary given by $\partial ( s_{i}(x) )$. We then compute

\begin{eqnarray*} y & = & \partial (s_ i(x) ) \\ & = & \sum _{j = 0}^{n+1} (-1)^{j} (d_ j \circ s_ i)(x) \\ & = & (\sum _{j = 0}^{i-1} (-1)^{j} (s_{i-1} \circ d_ j)(x) ) + (-1)^{j} x + (-1)^{j+1} x + (\sum _{j = i+2}^{n} (s_{i} \circ d_{j-1})(x) ) \\ & = & s_{i-1}( \sum _{j=0}^{i-1} (-1)^{j} d_ j(x) ) \\ & = & s_{i-1}( (\sum _{j=0}^{i-1} (-1)^{j} d_ j(x) ) + (\sum _{j=i+1}^{n} (-1)^{j} d_ j(x) ) ) \\ & = & s_{i-1}( \partial (x) - (-1)^{i} d_ i(x) ) \\ & = & (-1)^{i+1} (s_{i-1} \circ d_ i)(x). \end{eqnarray*}

Set $x' = x + (-1)^{i} y$. For $j \geq i$ we compute

\begin{eqnarray*} d_ j(x') & = & d_ j(x) + (-1)^{i} d_ j(y) \\ & = & d_ j(x) - (d_ j \circ s_{i-1} \circ d_ i)(x) \\ & = & \begin{cases} d_ j(x) - d_ i(x) & \text{ if } j = i \\ d_ j(x) - (s_{i-1} \circ d_ i \circ d_ j)(x) & \text{ if } j > i \end{cases}\\ & = & 0. \end{eqnarray*}

Our inductive hypothesis then guarantees that $x'$ is homologous to an element of the subgroup $\widetilde{\mathrm{N}}_{n}(A)$. Since $x$ is homologous to $x'$, it follows that $x$ is also homologous to an element of the subgroup $\widetilde{\mathrm{N}}_{n}(A)$. $\square$

Warning 2.5.6.23. Let $A_{\bullet }$ be a semisimplicial abelian group. Then we can still apply Construction 2.5.6.16 to define a subcomplex $\widetilde{ \mathrm{N} }_{\ast }(A)$ of the Moore complex $\mathrm{C}_{\ast }(A)$ (note that the definition of $\widetilde{ \mathrm{N} }_{\ast }(A)$ refers only to the face maps of $A_{\bullet }$). However, it is generally not true that the inclusion map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A)$ induces an isomorphism on homology unless $A_{\bullet }$ can be promoted to a simplicial abelian group.

We now turn to the proof of the Dold-Kan correspondence. The main ingredient is the following consequence of Proposition 2.5.6.19:

Proposition 2.5.6.24. Let $M_{\ast }$ be a chain complex and let $v: \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast } )) \rightarrow M_{\ast }$ be the counit map of Notation 2.5.6.11. Then:

• The map $v_0: \mathrm{N}_0( \mathrm{K}( M_{\ast } ) ) \rightarrow M_0$ is a monomorphism, whose image is the set $\mathrm{Z}_0(M)$ of $0$-cycles in $M_{\ast }$.

• For $n > 0$, the map $v_ n: \mathrm{N}_{n}( \mathrm{K}(M_{\ast } )) \rightarrow M_{n}$ is an isomorphism.

Proof. The first assertion follows from Example 2.5.6.5. To prove the second, fix $n > 0$ and let $f$ denote the composite map

$\widetilde{ \mathrm{N} }_{n}( \mathrm{K}(M_{\ast }) ) \hookrightarrow \mathrm{C}_{n}( \mathrm{K}( M_{\ast } ) ) \twoheadrightarrow \mathrm{N}_{n}( \mathrm{K}(M_{\ast }) ) \xrightarrow {v_ n} M_{n}.$

By virtue of Proposition 2.5.6.19, it will suffice to show that $f$ is an isomorphism. By definition, we can identify $\mathrm{C}_{n}( \mathrm{K}( M_{\ast } ) ) = \mathrm{K}_{n}( M_{\ast } )$ with the set of all chain maps $\sigma : \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow M_{\ast }$. Unwinding the definitions, we see that $\sigma$ belongs to the subgroup $\widetilde{ \mathrm{N} }_{n}( \mathrm{K}(M_{\ast }) ) \subseteq \mathrm{C}_{n}( \mathrm{K}( M_{\ast } ) )$ if and only if it annihilates the subcomplex $\mathrm{N}_{\ast }( \Lambda ^ n_0; \operatorname{\mathbf{Z}})$, where $\Lambda ^ n_0 \subset \Delta ^ n$ is the $0$-horn defined in Construction 1.1.2.9. We can therefore identify $\widetilde{ \mathrm{N} }_{n}( \mathrm{K}(M_{\ast }) )$ with the abelian group $\operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( K_{\ast }, M_{\ast } )$, where $K_{\ast }$ denotes the quotient of $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ by the subcomplex $\mathrm{N}_{\ast }( \Lambda ^ n_0; \operatorname{\mathbf{Z}})$. Note that there are exactly two nondegenerate simplices of $\Delta ^ n$ which do not belong to $\Lambda ^ n_0$; let us denote them by $\tau$ and $\tau '$ (where $\tau$ is of dimension $n$ and $\tau '$ of dimension $n-1$). Moreover, the differential on $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ satisfies $\partial (\tau ) \equiv \tau ' \pmod{ \mathrm{N}_{\ast }( \Lambda ^ n_ i; \operatorname{\mathbf{Z}}) }$. We conclude by observing that, under the preceding identification, the homomorphism $f: \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( K_{\ast }, M_{\ast } ) \rightarrow M_ n$ is given by evaluation on $\tau$, and is therefore an isomorphism. $\square$

Proof of Theorem 2.5.6.1. By virtue of Corollary 2.5.6.13, it will suffice to show that the construction $M_{\ast } \mapsto \mathrm{K}(M_{\ast })$ induces an equivalence of categories $\mathrm{K}: \operatorname{Ch}(\operatorname{\mathbf{Z}})_{\geq 0} \rightarrow \operatorname{ Ab }_{\Delta }$. We first show that the functor $\mathrm{K}$ is fully faithful when restricted to $\operatorname{Ch}(\operatorname{\mathbf{Z}})_{\geq 0}$. Let $M_{\ast }$ and $M'_{\ast }$ be chain complexes which are concentrated in degrees $\geq 0$; we wish to show that the canonical map

$\varphi : \operatorname{Hom}_{\operatorname{Ch}(\operatorname{\mathbf{Z}})}( M_{\ast }, M'_{\ast }) \rightarrow \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( \mathrm{K}( M_{\ast }), \mathrm{K}( M'_{\ast } ) )$

is an isomorphism. Let $\theta : \operatorname{Hom}_{\operatorname{ Ab }_{\Delta }}( \mathrm{K}( M_{\ast }), \mathrm{K}( M'_{\ast } ) ) \simeq \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}}) }( \mathrm{N}_{\ast }( \mathrm{K}( M_{\ast }) ), M'_{\ast } )$ be the isomorphism of Proposition 2.5.6.12. Unwinding the definitions, we see that $\theta \circ \varphi$ is given by precomposition with the counit map $v: \mathrm{N}_{\ast }( \mathrm{K}( M_{\ast }) ) \rightarrow M_{\ast }$ of Notation 2.5.6.11, and is therefore an isomorphism by virtue of Proposition 2.5.6.24 (together with our assumption that $M_{\ast }$ is concentrated in degrees $\geq 0$). It follows that $\varphi$ is also an isomorphism, as desired.

We now prove that the functor $\mathrm{K}: \operatorname{Ch}(\operatorname{\mathbf{Z}})_{\geq 0} \rightarrow \operatorname{ Ab }_{\Delta }$ is essentially surjective. Let $A_{\bullet }$ be a simplicial abelian group and let $M_{\ast } = \mathrm{N}_{\ast }(A)$ be its normalized Moore complex. Then there is a unique map of simplicial abelian groups $u: A_{\bullet } \rightarrow \mathrm{K}( M_{\ast } )$ for which the isomorphism

$\theta : \operatorname{Hom}_{\operatorname{ Ab }_{\Delta }}( A_{\bullet }, \mathrm{K}(M_{\ast } ) ) \rightarrow \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), M_{\ast } )$

of Proposition 2.5.6.12 carries $u$ to the identity map $\operatorname{id}: \mathrm{N}_{\ast }(A) \rightarrow M_{\ast }$. By construction, the induced map of normalized Moore complexes $\mathrm{N}_{\ast }(u): \mathrm{N}_{\ast }(A) \rightarrow \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast }))$ is right inverse to the counit map $v: \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast })) \rightarrow M_{\ast }$, which is an isomorphism by virtue of Proposition 2.5.6.24. Combining this observation with Proposition 2.5.6.19, we deduce that $u$ induces an isomorphism of chain complexes $\widetilde{\mathrm{N}}_{\ast }( A ) \rightarrow \widetilde{ \mathrm{N}}_{\ast }( \mathrm{K}(M_{\ast }) )$, and is therefore an isomorphism by virtue of Lemma 2.5.6.17. It follows that $A_{\bullet } \simeq \mathrm{K}(M_{\ast })$ belongs to the essential image of the functor $\mathrm{K}$, as desired. $\square$