
14.24 Dold-Kan

Lemma 14.24.1. Let $\mathcal{A}$ be an abelian category. The functor $N$ is faithful, and reflects isomorphisms, injections and surjections.

Proof. The faithfulness is immediate from the canonical splitting of Lemma 14.18.6. The statement on reflecting injections, surjections, and isomorphisms follows from Lemma 14.18.7. $\square$

Lemma 14.24.2. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $N : \mathcal{A} \to \mathcal{B}$, and $S : \mathcal{B} \to \mathcal{A}$ be functors. Suppose that

1. the functors $S$ and $N$ are exact,

2. there is an isomorphism $g : N \circ S \to \text{id}_\mathcal {B}$ to the identity functor of $\mathcal{B}$,

3. $N$ is faithful, and

4. $S$ is essentially surjective.

Then $S$ and $N$ are quasi-inverse equivalences of categories.

Proof. It suffices to construct a functorial isomorphism $S(N(A)) \cong A$. To do this choose $B$ and an isomorphism $f : A \to S(B)$. Consider the map

$f^{-1} \circ g_{S(B)} \circ S(N(f)) : S(N(A)) \to S(N(S(B))) \to S(B) \to A.$

It is easy to show this does not depend on the choice of $f, B$ and gives the desired isomorphism $S \circ N \to \text{id}_\mathcal {A}$. $\square$

Theorem 14.24.3. Let $\mathcal{A}$ be an abelian category. The functor $N$ induces an equivalence of categories

$N : \text{Simp}(\mathcal{A}) \longrightarrow \text{Ch}_{\geq 0}(\mathcal{A})$

Proof. We will describe a functor in the reverse direction inspired by the construction of Lemma 14.22.4 (except that we throw in a sign to get the boundaries right). Let $A_\bullet$ be a chain complex with boundary maps $d_{A, n} : A_ n \to A_{n - 1}$. For each $n \geq 0$ denote

$I_ n = \Big\{ \alpha : [n] \to \{ 0, 1, 2, \ldots \} \mid \mathop{\mathrm{Im}}(\alpha ) = [k]\text{ for some }k \Big\} .$

For $\alpha \in I_ n$ we denote $k(\alpha )$ the unique integer such that $\mathop{\mathrm{Im}}(\alpha ) = [k]$. We define a simplicial object $S(A_\bullet )$ as follows:

1. $S(A_\bullet )_ n = \bigoplus _{\alpha \in I_ n} A_{k(\alpha )}$, which we will write as $\bigoplus _{\alpha \in I_ n} A_{k(\alpha )} \cdot \alpha$ to suggest thinking of “$\alpha$” as a basis vector for the summand corresponding to it,

2. given $\varphi : [m] \to [n]$ we define $S(A_\bullet )(\varphi )$ by its restriction to the direct summand $A_{k(\alpha )} \cdot \alpha$ of $S(A_\bullet )_ n$ as follows

1. $\alpha \circ \varphi \not\in I_ m$ then we set it equal to zero,

2. $\alpha \circ \varphi \in I_ m$ but $k(\alpha \circ \varphi )$ not equal to either $k(\alpha )$ or $k(\alpha ) - 1$ then we set it equal to zero as well,

3. if $\alpha \circ \varphi \in I_ m$ and $k(\alpha \circ \varphi ) = k(\alpha )$ then we use the identity map to the summand $A_{k(\alpha \circ \varphi )} \cdot (\alpha \circ \varphi )$ of $S(A_\bullet )_ m$, and

4. if $\alpha \circ \varphi \in I_ m$ and $k(\alpha \circ \varphi ) = k(\alpha ) - 1$ then we use $(-1)^{k(\alpha )} d_{A, k(\alpha )}$ to the summand $A_{k(\alpha \circ \varphi )}\cdot (\alpha \circ \varphi )$ of $S(A_\bullet )_ m$.

It is an exercise (FIXME) to show that this is a simplicial complex; one has to use in particular that the compositions $d_{A, k} \circ d_{A, k - 1}$ are all zero.

Having verified this, the correct way to proceed with the proof would be to prove directly that $N$ and $S$ are quasi-inverse functors (FIXME). Instead we prove this by an indirect method using Eilenberg-Maclane objects and truncations. It is clear that $A_\bullet \mapsto S(A_\bullet )$ is an exact functor from chain complexes to simplicial objects. If $A_ i = 0$ for $i = 0, \ldots , n$ then $S(A_\bullet )_ i = 0$ for $i = 0, \ldots , n$. The objects $K(A, k)$, see Definition 14.22.3, are equal to $S(A[-k])$ where $A[-k]$ is the chain complex with $A$ in degree $k$ and zero elsewhere.

Moreover, for each integer $k$ we get a sub simplicial object $S_{\leq k}(A_\bullet )$ by considering only those $\alpha$ with $k(\alpha ) \leq k$. In fact this is nothing but $S(\sigma _{\leq k}A_\bullet )$, where $\sigma _{\leq k}A_\bullet$ is the “stupid” truncation of $A_\bullet$ at $k$ (which simply replaces $A_ i$ by $0$ for $i > k$). Also, by Lemma 14.21.10 we see that it is equal to $i_{k!}\text{sk}_ k S(A_\bullet )$. Clearly, the quotient $S_{\leq k}(A_\bullet )/S_{\leq k - 1}(A_\bullet ) = K(A_ k, k)$ and the quotient $S(A_\bullet )/S_{\leq k}(A_\bullet ) = S(A/\sigma _{\leq k}A_\bullet )$ is a simplicial object whose $i$th term is zero for $i = 0, \ldots , k$. Since $S_{\leq k - 1}(A_\bullet )$ is filtered with subquotients $K(A_ i, i)$, $i < k$ we see that $N(S_{\leq k - 1}(A_\bullet ))_ k = 0$ by exactness of the functor $N$, see Lemma 14.23.7. All in all we conclude that the maps

$N(S(A_\bullet ))_ k \leftarrow N(S_{\leq k}(A_\bullet ))_ k \to N(S(A_ k[-k])) = N(K(A_ k, k))_ k = A_ k$

are functorial isomorphisms.

It is actually easy to identify the map $A_ k \to N(S(A_\bullet ))_ k$. Note that there is a unique map $A_ k \to S(A_\bullet )_ k$ corresponding to the summand $\alpha = \text{id}_{[k]}$. Note that $\mathop{\mathrm{Im}}(\text{id}_{[k]} \circ \delta ^ k_ i)$ has cardinality $k - 1$ but does not have image $[k - 1]$ unless $i = k$. Hence $d^ k_ i$ kills the summand $A_ k \cdot \text{id}_{[k]}$ for $i = 0, \ldots , k - 1$. From the abstract computation of $N(S(A_\bullet ))_ k$ above we conclude that the summand $A_ k \cdot \text{id}_{[k]}$ is equal to $N(S(A_\bullet ))_ k$.

In order to show that $N \circ S$ is the identity functor on $\text{Ch}_{\geq 0}(\mathcal{A})$, the last thing we have to verify is that we recover the map $d_{A, k + 1} : A_{k + 1} \to A_ k$ as the differential on the complex $N(S(A_\bullet ))$ as follows

$A_{k + 1} = N(S(A_\bullet ))_{k + 1} \to N(S(A_\bullet ))_ k = A_ k$

By definition the map $N(S(A_\bullet ))_{k + 1} \to N(S(A_\bullet ))_ k$ corresponds to the restriction of $(-1)^{k + 1}d^{k + 1}_{k + 1}$ to $N(S(A_\bullet ))$ which is the summand $A_{k + 1} \cdot \text{id}_{[k + 1]}$. And by the definition of $S(A_\bullet )$ above the map $d^{k + 1}_{k + 1}$ maps $A_{k + 1} \cdot \text{id}_{[k + 1]}$ into $A_ k \cdot \text{id}_{[k]}$ by $(-1)^{k + 1}d_{A, k + 1}$. The signs cancel and hence the desired equality.

We know that $N$ is faithful, see Lemma 14.24.1. If we can show that $S$ is essentially surjective, then it will follow that $N$ is an equivalence, see Homology, Lemma 14.24.2. Note that if $A_\bullet$ is a chain complex then $S(A_\bullet ) = \mathop{\mathrm{colim}}\nolimits _ n S_{\leq n}(A_\bullet ) = \mathop{\mathrm{colim}}\nolimits _ n S(\sigma _{\leq n} A_\bullet ) = \mathop{\mathrm{colim}}\nolimits _ n i_{n!} \text{sk}_ n S(A_\bullet )$ by construction of $S$. By Lemma 14.22.5 it suffices to show that $i_{n!} V$ is in the essential image for any $n$-truncated simplicial object $V$. By induction on $n$ it suffices to show that any extension

$0 \to S(A_\bullet ) \to V \to K(A, n) \to 0$

where $A_ i = 0$ for $i \geq n$ is in the essential image of $S$. By Homology, Lemma 12.7.3 we have abelian group homomorphisms

$\xymatrix{ \mathop{\mathrm{Ext}}\nolimits _{\text{Simp}(\mathcal{A})}(K(A, n), S(A_\bullet )) \ar@<1ex>[r]^-{N} & \mathop{\mathrm{Ext}}\nolimits _{\text{Ch}_{\geq 0}(\mathcal{A})}(A[-n], A_\bullet ) \ar@<1ex>[l]^-{S} }$

between ext groups (see Homology, Definition 12.6.2). We want to show that $S$ is surjective. We know that $N \circ S = \text{id}$. Hence it suffices to show that $\mathop{\mathrm{Ker}}(N) = 0$. Clearly an extension

$\xymatrix{ & 0 \ar[r] & 0 \ar[r] \ar[d] & A_{n - 1} \ar[r] \ar[d] & A_{n - 2} \ar[r] \ar[d] & \ldots \ar[r] & A_0 \ar[r] \ar[d] & 0 \\ E : & 0 \ar[r] & A \ar[r] \ar[d] & A_{n - 1} \ar[r] \ar[d] & A_{n - 2} \ar[r] \ar[d] & \ldots \ar[r] & A_0 \ar[r] \ar[d] & 0 \\ & 0 \ar[r] & A \ar[r] & 0 \ar[r] & 0 \ar[r] & \ldots \ar[r] & 0 \ar[r] & 0 }$

of $A_\bullet$ by $A[-n]$ in $\text{Ch}(\mathcal{A})$ is zero if and only if the map $A \to A_{n - 1}$ is zero. Thus we have to show that any extension

$0 \to S(A_\bullet ) \to V \to K(A, n) \to 0$

such that $A = N(V)_ n \to N(V)_{n - 1}$ is zero is split. By Lemma 14.22.2 we have

$\mathop{Mor}\nolimits (K(A, n), V) = \left\{ f : A \to \bigcap \nolimits _{i = 0}^ n \mathop{\mathrm{Ker}}(d^ n_ i : V_ n \to V_{n - 1}) \right\}$

and if $A = N(V)_ n \to N(V)_{n - 1}$ is zero, then the intersection occurring in the formula above is equal to $A$. Let $i : K(A, n) \to V$ be the morphism corresponding to $\text{id}_ A$ on the right hand side of the displayed formula. Clearly this is a section to the map $V \to K(A, n)$ and the extension is split as desired. $\square$

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