Proposition 2.5.6.23. Let $M_{\ast }$ be a chain complex and let $v: \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast } )) \rightarrow M_{\ast }$ be the counit map of Notation 2.5.6.10. Then:
The map $v_0: \mathrm{N}_0( \mathrm{K}( M_{\ast } ) ) \rightarrow M_0$ is a monomorphism, whose image is the set $\mathrm{Z}_0(M)$ of $0$-cycles in $M_{\ast }$.
For $n > 0$, the map $v_ n: \mathrm{N}_{n}( \mathrm{K}(M_{\ast } )) \rightarrow M_{n}$ is an isomorphism.
Proof. The first assertion follows from Example 2.5.6.5. To prove the second, fix $n > 0$ and let $f$ denote the composite map
By virtue of Proposition 2.5.6.17, it will suffice to show that $f$ is an isomorphism. By definition, we can identify $\mathrm{C}_{n}( \mathrm{K}( M_{\ast } ) ) = \mathrm{K}_{n}( M_{\ast } )$ with the set of all chain maps $\sigma : \mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow M_{\ast }$. Unwinding the definitions, we see that $\sigma $ belongs to the subgroup $\widetilde{ \mathrm{N} }_{n}( \mathrm{K}(M_{\ast }) ) \subseteq \mathrm{C}_{n}( \mathrm{K}( M_{\ast } ) )$ if and only if it annihilates the subcomplex $\mathrm{N}_{\ast }( \Lambda ^ n_0; \operatorname{\mathbf{Z}})$, where $\Lambda ^ n_0 \subset \Delta ^ n$ is the $0$-horn defined in Construction 1.2.4.1. We can therefore identify $\widetilde{ \mathrm{N} }_{n}( \mathrm{K}(M_{\ast }) )$ with the abelian group $\operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( K_{\ast }, M_{\ast } )$, where $K_{\ast }$ denotes the quotient of $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ by the subcomplex $\mathrm{N}_{\ast }( \Lambda ^ n_0; \operatorname{\mathbf{Z}})$. Note that there are exactly two nondegenerate simplices of $\Delta ^ n$ which do not belong to $\Lambda ^ n_0$; let us denote them by $\tau $ and $\tau '$ (where $\tau $ is of dimension $n$ and $\tau '$ of dimension $n-1$). Moreover, the differential on $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}})$ satisfies $\partial (\tau ) \equiv \tau ' \pmod{ \mathrm{N}_{\ast }( \Lambda ^ n_0; \operatorname{\mathbf{Z}}) }$. We conclude by observing that, under the preceding identification, the homomorphism $f: \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( K_{\ast }, M_{\ast } ) \rightarrow M_ n$ is given by evaluation on $\tau $, and is therefore an isomorphism. $\square$