# Kerodon

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Proposition 2.5.6.12. 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 2.5.6.11. Then, for any simplicial abelian group $A_{\bullet }$, the composite map

$\theta : \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( A_{\bullet }, \mathrm{K}( M_{\ast }) ) \rightarrow \operatorname{Hom}_{\operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), \mathrm{N}_{\ast }( \mathrm{K}( M_{\ast }) ) ) \xrightarrow {v \circ } \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), M_{\ast } )$

is an isomorphism of abelian groups.

Proof. Let us say that a simplicial abelian group $A_{\bullet }$ is free if it can be written as a (possibly infinite) direct sum of simplicial abelian groups of the form $\operatorname{\mathbf{Z}}[ \Delta ^ n ]$. Note that every simplicial abelian group $A_{\bullet }$ admits a surjection $P_{\bullet } \twoheadrightarrow A_{\bullet }$, where $P_{\bullet }$ is free (for example, we can take $P_{\bullet }$ to be the direct sum $\bigoplus _{\sigma } \operatorname{\mathbf{Z}}[ \Delta ^{ \mathrm{dim}(\sigma ) }]$ where $\sigma$ ranges over all the simplices of $A_{\bullet }$). Applying this observation twice, we observe that every simplicial abelian group $A_{\bullet }$ admits a resolution

$Q_{\bullet } \rightarrow P_{\bullet } \twoheadrightarrow A_{\bullet } \rightarrow 0,$

which determines a commutative diagram of exact sequences

$\xymatrix@R =50pt@C=25pt{0 \ar [r] & \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( A_{\bullet }, \mathrm{K}( M_{\ast }) ) \ar [r] \ar [d]^{\theta } & \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( P_{\bullet }, \mathrm{K}( M_{\ast }) ) \ar [d]^{\theta '} \ar [r] & \operatorname{Hom}_{ \operatorname{ Ab }_{\Delta }}( Q_{\bullet }, \mathrm{K}( M_{\ast }) ) \ar [d]^{\theta ''} \\ 0 \ar [r] & \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(A), M_{\ast } ) \ar [r] & \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(P), M_{\ast } ) \ar [r] & \operatorname{Hom}_{ \operatorname{Ch}(\operatorname{\mathbf{Z}})}( \mathrm{N}_{\ast }(Q), M_{\ast } ). }$

Consequently, to prove that $\theta$ is an isomorphism, it will suffice to show that $\theta '$ and $\theta ''$ are isomorphisms. In other words, we may assume without loss of generality that the simplicial abelian group $A_{\bullet }$ is free. Decomposing $A_{\bullet }$ as a direct sum, we can further reduce to the case $A_{\bullet } = \operatorname{\mathbf{Z}}[ \Delta ^ n ]$, in which case the result follows immediately from the definitions. $\square$