Lemma 14.24.1. Let $\mathcal{A}$ be an abelian category. The functor $N$ is faithful, and reflects isomorphisms, injections and surjections.
14.24 Dold-Kan
In this section we prove the Dold-Kan theorem relating simplicial objects in an abelian category with chain complexes.
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
the functors $S$ and $N$ are exact,
there is an isomorphism $g : N \circ S \to \text{id}_\mathcal {B}$ to the identity functor of $\mathcal{B}$,
$N$ is faithful, and
$S$ 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
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
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
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:
$S(A_\bullet )_ n = \bigoplus _{\alpha \in I_ n} A_{k(\alpha )}$, 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,
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
$\alpha \circ \varphi \not\in I_ m$ then we set it equal to zero,
$\alpha \circ \varphi \in I_ m$ but $k(\alpha \circ \varphi )$ not equal to either $k(\alpha )$ or $k(\alpha ) - 1$ then we set it equal to zero as well,
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
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 $S(A_\bullet )_ m$.
Let us show that $S(A_\bullet )$ is a simplicial object of $\mathcal{A}$. To do this, assume we have maps $\varphi : [m] \to [n]$ and $\psi : [n] \to [p]$. We will show that $S(A_\bullet )(\varphi ) \circ S(A_\bullet )(\psi ) = S(A_\bullet )(\psi \circ \varphi )$. Choose $\beta \in I_ p$ and set $\alpha = \beta \circ \psi $ and $\gamma = \alpha \circ \varphi $ viewed as maps $\alpha : [n] \to \{ 0, 1, 2, \ldots \} $ and $\gamma : [m] \to \{ 0, 1, 2, \ldots \} $. Picture
We will show that the restriction of the maps $S(A_\bullet )(\varphi ) \circ S(A_\bullet )(\psi )$ and $S(A_\bullet )(\psi \circ \varphi )$. to the summand $A_{k(\beta )} \cdot \beta $ agree. There are several cases to consider
Say $\alpha \not\in I_ n$ so the restriction of $S(A_\bullet )(\psi )$ to $A_{k(\beta )} \cdot \beta $ is zero. Then either $\gamma \not\in I_ m$ or we have $[k(\gamma )] = \mathop{\mathrm{Im}}(\gamma ) \subset \mathop{\mathrm{Im}}(\alpha ) \subset [k(\beta )]$ and the subset $\mathop{\mathrm{Im}}(\alpha )$ of $[k(\beta )]$ has a gap so $k(\gamma ) < k(\beta ) - 1$. In both cases we see that the restriction of $S(A_\bullet )(\psi \circ \varphi )$ to $A_{k(\beta )} \cdot \beta $ is zero as well.
Say $\alpha \in I_ n$ and $k(\alpha ) < k(\beta ) - 1$ so the restriction of $S(A_\bullet )(\psi )$ to $A_{k(\beta )} \cdot \beta $ is zero. Then either $\gamma \not\in I_ m$ or we have $[k(\gamma )] \subset [k(\alpha )] \subset [k(\beta )]$ and it follows that $k(\gamma ) < k(\beta ) - 1$. In both cases we see that the restriction of $S(A_\bullet )(\psi \circ \varphi )$ to $A_{k(\beta )} \cdot \beta $ is zero as well.
Say $\alpha \in I_ n$ and $k(\alpha ) = k(\beta )$ so the restriction of $S(A_\bullet )(\psi )$ to $A_{k(\beta )} \cdot \beta $ is the identity map from $A_{k(\beta )} \cdot \beta $ to $A_{k(\alpha )} \cdot \alpha $. In this case because $\mathop{\mathrm{Im}}(\alpha ) =[k(\beta )]$ the rule describing the restriction of $S(A_\bullet )(\psi \circ \varphi )$ to the summand $A_{k(\beta )} \cdot \beta $ is exactly the same as the rule describing the restriction of $S(A_\bullet )(\varphi )$ to the summand $A_{k(\alpha )} \cdot \alpha $ and hence agreement holds.
Say $\alpha \in I_ n$ and $k(\alpha ) = k(\beta ) - 1$ so the restriction of $S(A_\bullet )(\psi )$ to $A_{k(\beta )} \cdot \beta $ is given by $(-1)^{k(\beta )}d_{A, k(\beta )}$ to $A_{k(\alpha )} \cdot \alpha $. Subcases
If $\gamma \not\in I_ m$, then both the restriction of $S(A_\bullet )(\psi \circ \varphi )$ to the summand $A_{k(\beta )} \cdot \beta $ and the restriction of $S(A_\bullet )(\varphi )$ to the summand $A_{k(\alpha )} \cdot \alpha $ are zero and we get agreement.
If $\gamma \in I_ m$ but $k(\gamma ) < k(\alpha ) - 1$, then again both restrictions are zero and we get agreement.
If $\gamma \in I_ m$ and $k(\gamma ) = k(\alpha )$ then $\mathop{\mathrm{Im}}(\gamma ) = \mathop{\mathrm{Im}}(\alpha )$. In this case the restriction of $S(A_\bullet )(\psi \circ \varphi )$ to the summand $A_{k(\beta )} \cdot \beta $ is given by $(-1)^{k(\beta )}d_{A, k(\beta )}$ to $A_{k(\gamma )} \cdot \gamma $ and the restriction of $S(A_\bullet )(\varphi )$ to the summand $A_{k(\alpha )} \cdot \alpha $ is the identity map $A_{k(\alpha )} \cdot \alpha \to A_{k(\gamma )} \cdot \gamma $. Hence agreement holds.
Finally, if $\gamma \in I_ m$ and $k(\gamma ) = k(\alpha ) - 1$ then the restriction of $S(A_\bullet )(\varphi )$ to the summand $A_{k(\alpha )} \cdot \alpha $ is given by $(-1)^{k(\alpha )} d_{A, k(\alpha )}$ as a map $A_{k(\alpha )} \cdot \alpha \to A_{k(\beta )} \cdot \beta $. Since $A_\bullet $ is a complex we see that the composition $A_{k(\beta )} \cdot \beta \to A_{k(\alpha )} \cdot \alpha \to A_{k(\gamma )} \cdot \gamma $ is zero which matches what we get for the restriction of $S(A_\bullet )(\psi \circ \varphi )$ to the summand $A_{k(\beta )} \cdot \beta $ because $k(\gamma ) = k(\beta ) - 2 < k(\beta ) - 1$.
Thus $S(A_\bullet )$ is a simplicial object of $\mathcal{A}$.
Let us construct an isomorphism $A_\bullet \to N(S(A_\bullet ))$ functorial in $A_\bullet $. Recall that
as chain complexes by Lemma 14.23.6. On the other hand it follows from Remark 14.23.7 and the construction of $S(A_\bullet )$ that
However, if $\alpha \in I_ n$ then we have $k(\alpha ) \geq n \Leftrightarrow \alpha = \text{id}_{[n]} : [n] \to [n]$. Thus the summand $A_ n \cdot \text{id}_{[n]}$ of $S(A_\bullet )_ n$ is a complement to the summand $D(S(A_\bullet ))_ n$. All the maps $d^ n_ i : S(A_\bullet )_ n \to S(A_\bullet )_ n$ restrict to zero on the summand $A_ n \cdot \text{id}_{[n]}$ except for $d^ n_ n$ which produces $(-1)^ n d_{A, n}$ from $A_ n \cdot \text{id}_{[n]}$ to $A_{n - 1} \cdot \text{id}_{[n - 1]}$. We conclude that $A_ n \cdot \text{id}_{[n]}$ must be equal to the summand $N(S(A_\bullet ))_ n$ and moreover the restriction of the differential $d_ n = \sum (-1)^ id^ n_ i : S(A_\bullet )_ n \to S(A_\bullet )_{n - 1}$ to the summand $A_ n \cdot \text{id}_{[n]}$ gives what we want!
Finally, we have to show that $S \circ N$ is isomorphic to the identity functor. Let $U$ be a simplicial object of $\mathcal{A}$. Then we can define an obvious map
by using $U(\alpha ) : N(U)_{k(\alpha )} \to U_ n$ on the summand corresponding to $\alpha $. By Definition 14.18.1 this is an isomorphism. To finish the proof we have to show that this is compatible with the maps in the simplicial objects. Thus let $\varphi : [m] \to [n]$ and let $\alpha \in I_ n$. Set $\beta = \alpha \circ \varphi $. Picture
There are several cases to consider
Say $\beta \not\in I_ m$. Then there exists an index $0 \leq j < k(\alpha )$ with $j \not\in \mathop{\mathrm{Im}}(\alpha \circ \varphi )$ and hence we can choose a factorization $\alpha \circ \varphi = \delta ^{k(\alpha )}_ j \circ \psi $ for some $\psi : [m] \to [k(\alpha ) - 1]$. It follows that $U(\varphi )$ is zero on the image of the summand $N(U)_{k(\alpha )} \cdot \alpha $ because $U(\varphi ) \circ U(\alpha ) = U(\alpha \circ \varphi ) = U(\psi ) \circ d^{k(\alpha )}_ j$ is zero on $N(U)_{k(\alpha )}$ by construction of $N$. This matches our rule for $S(N(U))$ given above.
Say $\beta \in I_ m$ and $k(\beta ) < k(\alpha ) - 1$. Here we argue exactly as in case (1) with $j = k(\alpha ) - 1$.
Say $\beta \in I_ m$ and $k(\beta ) = k(\alpha )$. Here the summand $N(U)_{k(\alpha )} \cdot \alpha $ is mapped by the identity to the summand $N(U)_{k(\beta )} \cdot \beta $. This is the same as the effect of $U(\varphi )$ since in this case $U(\varphi ) \circ U(\alpha ) = U(\beta )$.
Say $\beta \in I_ m$ and $k(\beta ) = k(\alpha ) - 1$. Here we use the differential $(-1)^{k(\alpha )} d_{N(U), k(\alpha )}$ to map the summand $N(U)_{k(\alpha )} \cdot \alpha $ to the summand $N(U)_{k(\beta )} \cdot \beta $. On the other hand, since $\mathop{\mathrm{Im}}(\beta ) = [k(\beta )]$ in this case we get $\alpha \circ \varphi = \delta ^{k(\alpha )}_{k(\alpha )} \circ \beta $. Thus we see that $U(\varphi )$ composed with the restriction of $U(\alpha )$ to $N(U)_{k(\alpha )}$ is equal to $U(\beta )$ precomposed with $d^{k(\alpha )}_{k(\alpha )}$ restricted to $N(U)_{k(\alpha )}$. Since $d_{N(U), k(\alpha )} = \sum (-1)^ i d^{k(\alpha )}_ i$ and since $d^{k(\alpha )}_ i$ restricts to zero on $N(U)_{k(\alpha )}$ for $i < k(\alpha )$ we see that equality holds.
This finishes the proof of the theorem. $\square$
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