Kerodon

Proof of Theorem 2.5.6.1. By virtue of Corollary 2.5.6.12, 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

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 2.5.6.11. 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 2.5.6.10, and is therefore an isomorphism by virtue of Proposition 2.5.6.23 (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

of Proposition 2.5.6.11 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 2.5.6.23. Combining this observation with Proposition 2.5.6.17, 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 2.5.6.16. It follows that $A_{\bullet } \simeq \mathrm{K}(M_{\ast })$ belongs to the essential image of the functor $\mathrm{K}$, as desired. $\square$