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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