Consider the Moore complex functor $M : \mathsf{sAb} \to \mathrm{Ch}^+(\mathsf{Ab})$, where the complex associated to a simplicial abelian group $A$ is $MA_n = A_n$ with boundary $\partial = \sum_{j=0}^n (-1)^j d_j$. If we look at the normalized subcomplex $NA_n = \ker d_0 \cap \ldots \cap \ker d_{n-1}$ then we get another functor $N : \mathsf{sAb} \to \mathrm{Ch}^+(\mathsf{Ab})$, which the Dold-Kan Correspondence says is an equivalence of categories (and further a Quillen equivalence). The fact that we need to normalize here makes me think $M$ itself is not an equivalence $\mathsf{sAb} \to \mathrm{Ch}^+(\mathsf{Ab})$, but I don't see how to prove it. I'm also wondering whether $M$ even has an adjoint (on either side). It seems like $M$ preserves limits, since limits in both $\mathsf{sAb}$ and $\mathrm{Ch}^+(\mathsf{Ab})$ are computed degreewise, and $(MA)_n = A_n$.

Edit: I checked that $M$ preserves all limits and colimits, and according to the nlab $\mathsf{sAb}$ is a "total" and "cototal" category (as it's the category of abelian sheaves on a site) so an adjoint functor theorem ensures $M$ has adjoints on both sides. What are these adjoints? Do they have a nice interpretation?

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    $\begingroup$ I've deleted my answer; you are absolutely right: I was thinking of the Moore complex as a functor from the category of simplicial sets, not groups. In that case indeed the image of $\Delta^0$ is not the zero complex. $\endgroup$
    – qualcuno
    Mar 3, 2021 at 23:19
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    $\begingroup$ To see that $M$ is not an equivalence I think it suffices to observe that $M$ sends any non-zero simplicial abelian group to a chain complex which is non-zero in any positive degree (because of the degeneracies). Therefore, no chain complex in the image of $M$ can be isomorphic to a chain complex concentrated in one degree. Thus $M$ is not essentially surjective and not an equivalence. $\endgroup$
    – Charlie
    Mar 4, 2021 at 10:29

1 Answer 1


By this result claimed on the nlab the free forgetful adjunction $F \dashv U$ between abelian groups and sets gives rise to a biadjunction(?) $\overline{F} \dashv \overline{U}$ between the bicategories $\mathsf{Ab}-\mathsf{Cat}$ and $(\mathsf{Set}-)\mathsf{Cat}$. My understanding is that $\overline{F}$ takes an ordinary category $\mathsf{C}$ and constructs an $\mathsf{Ab}$-enriched category $\overline{F}(\mathsf{C})$ which has the same objects as $\mathsf{C}$ but $\operatorname{Hom}_{\overline{F}(\mathsf{C})}(x, y) = F(\operatorname{Hom}_{\mathsf{C}}(x, y))$ whereas $\overline{U}$ takes an $\mathsf{Ab}$-enriched category $\mathcal{S}$ and constructs an ordinary category $\overline{U}(\mathcal{S})$ which has the same objects as $\mathcal{S}$ but $\operatorname{Hom}_{\overline{U}(\mathcal{S})}(x, y) = U(\operatorname{Hom}_{\mathcal{S}}(x, y))$ and we have pseudonatural adjoint equivalences $\mathsf{Ab-Fun}(\overline{F}(\mathsf{C}), \mathcal{S}) \simeq \operatorname{Fun}(\mathsf{C}, \overline{U}(\mathcal{S}))$.

In particular the $\mathsf{Ab}$-enriched category $+\Delta \stackrel{\text{def}}= \overline{F}(\Delta)$ has the structure of a mapping out property $\mathsf{Ab-Fun}(+\Delta^{\mathrm{op}}, \mathcal{M}) \simeq \operatorname{Fun}(\Delta^{\mathrm{op}}, \overline{U}(\mathcal{M})) = s\overline{U}(\mathcal{M})$. This is classified by a simplicial object of $+\Delta^{\mathrm{op}}$, dually a cosimplicial object of $+\Delta$, which can be calculated as the unit of the $2$-adjunction $\overline{F} \dashv \overline{U}$ at component $\Delta$, which is the identity on objects and the unit of $F \dashv U$ on $\operatorname{Hom}$-sets, i.e. the identity on objects and the inclusion of basis elements on morphisms. So the cosimplicial object $A$ of $+\Delta$ has $A_n = [n]$, and the coface maps $d^i : A_n \to A_{n-1}$ are defined on basis elements by $d^i(f) = \delta^i(f)$ and extended additively (similarly for codegeneracy maps).

Now because $\mathsf{Ab}$ is a nice enough enriching category (cosmos I think?) we can promote the ordinary category $\mathsf{Ab-Fun}(\mathcal{R}, \mathcal{S})$ to an $\mathsf{Ab}$-enriched category $+\mathsf{Fun}(\mathcal{R}, \mathcal{S})$ with $\overline{U}(+\mathsf{Fun}(\mathcal{R}, \mathcal{S})) = \mathsf{Ab-Fun}(\mathcal{R}, \mathcal{S})$. This is just "remembering" that we can add and subtract morphisms between abelian groups.

The category of enriched presheaves is $\mathsf{Mod}(\mathcal{R}) \stackrel{\text{def}}= +\mathsf{Fun}(\mathcal{R}^{\mathrm{op}}, \mathsf{Ab})$ is the free $\mathsf{Ab}$-cocompletion of $\mathcal{R}$. Note $\mathsf{Mod}(+\Delta) = s\mathsf{Ab}$.

Connective chain complexes of abelian groups are also the category of modules over an $\mathrm{Ab}$-enriched category $\mathcal{R}$. This category is $\mathcal{R} = \overline{F}(\mathbb{N}^{\leq})/J$, where $\mathbb{N}^{\leq}$ is the poset category of the nonnegative integers and $J$ is the ideal generated by all morphisms of the form $d_n^2$, where $d_n$ is the basis element corresponding to the element of $\mathrm{Hom}_{\mathbb{N}^{\leq}}(n, n + 1)$. So the subgroup of $\mathrm{Hom}_{\mathbb{N}^{\leq}}(n, m)$ given by $J$ is $0$ if $m = n + 1$ or $m = n$ and is the entire subgroup otherwise. More concretely $\mathcal{R}$ has objects the natural numbers and $$\operatorname{Hom}_{\mathcal{R}}(n, m) = \begin{cases} F(\{d\}) &\text{ if } m = n + 1 \\ F(\{\mathrm{id}_n\}) &\text{ if } m = n \\ 0 &\text{ otherwise.} \end{cases}$$

The dual of the standard proof that the differential of the Moore complex squares to $0$ proves that we have a well defined (additive) functor $M' : \mathcal{R} \to \Delta+$, sending $n$ to $[n]$ and sending the differential to the alternating sum of coface maps. Composing this with the enriched Yoneda embedding gives rise to an adjunction between the essentially unique $\mathsf{Ab}$-enriched colimit preserving extension of $y \circ M'$, a functor $\mathsf{Mod}(\mathcal{R}) = \mathsf{Ch}^+(\mathbb{Z}) \to s\mathsf{Ab}$, and the nerve of $y \circ M'$, a functor $s\mathsf{Ab} \to \mathsf{Ch}^+(\mathbb{Z})$. This nerve sends a simplicial abelian group to its alternating face map complex, i.e. is the Moore complex functor. This "explains" the left adjoint of the Moore complex functor. But also because the functor $y\circ M'$ is valued in projective objects (representables) the nerve $M$ automatically preserves colimits (one can see this from the defintion of the nerve and the fact that preservation of colimits can be checked after evaluating at each object). Hence it has a further right adjoint, although I still don't know a better direct description of this adjoint.

It's interesting to me that the Moore complex arises from a functor of small categories $\mathcal{R} \to +\Delta$ while the better behaved normalized Moore complex requires a cosimplicial object in the large category functor $\mathsf{Ch}^+(\mathbb{Z})$. This cosimplicial chain complex is in degree $n$ the the complex which in degree $m$ is the free abelian group on injections $[m] \hookrightarrow [n]$ and whose differential is the alternating sum of the precompositions with coface maps. The coface maps of this cosimplicial object postcompose with the coface maps in the simplex category and the codegeneracy maps are all zero. My two thoughts here are that we didn't actually need the whole category $\mathsf{Ch}^+(\mathbb{Z})$, the complexes involved are all perfect, and that I should think about the semisimplicial version of Dold Kan.


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