The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.70.5. Let $R$ be a ring. Let $M_1, M_2, N$ be $R$-modules. Suppose that $F_{\bullet }$ is a free resolution of the module $M_1$, and $G_{\bullet }$ is a free resolution of the module $M_2$. Let $\varphi : M_1 \to M_2$ be a module map. Let $\alpha : F_{\bullet } \to G_{\bullet }$ be a map of complexes inducing $\varphi $ on $M_1 = \mathop{\mathrm{Coker}}(d_{F, 1}) \to M_2 = \mathop{\mathrm{Coker}}(d_{G, 1})$, see Lemma 10.70.4. Then the induced maps

\[ H^ i(\alpha ) : H^ i(\mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N)) \longrightarrow H^ i(\mathop{\mathrm{Hom}}\nolimits _ R(G_{\bullet }, N)) \]

are independent of the choice of $\alpha $. If $\varphi $ is an isomorphism, so are all the maps $H^ i(\alpha )$. If $M_1 = M_2$, $F_\bullet = G_\bullet $, and $\varphi $ is the identity, so are all the maps $H_ i(\alpha )$.

Proof. Another map $\beta : F_{\bullet } \to G_{\bullet }$ inducing $\varphi $ is homotopic to $\alpha $ by Lemma 10.70.4. Hence the maps $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) \to \mathop{\mathrm{Hom}}\nolimits _ R(G_\bullet , N)$ are homotopic. Hence the independence result follows from Lemma 10.70.3.

Suppose that $\varphi $ is an isomorphism. Let $\psi : M_2 \to M_1$ be an inverse. Choose $\beta : G_{\bullet } \to F_{\bullet }$ be a map inducing $\psi : M_2 = \mathop{\mathrm{Coker}}(d_{G, 1}) \to M_1 = \mathop{\mathrm{Coker}}(d_{F, 1})$, see Lemma 10.70.4. OK, and now consider the map $H^ i(\alpha ) \circ H^ i(\beta ) = H^ i(\alpha \circ \beta )$. By the above the map $H^ i(\alpha \circ \beta )$ is the same as the map $H^ i(\text{id}_{G_{\bullet }}) = \text{id}$. Similarly for the composition $H^ i(\beta ) \circ H^ i(\alpha )$. Hence $H^ i(\alpha )$ and $H^ i(\beta )$ are inverses of each other. $\square$


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