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 10.70.4. Let $R$ be a ring. Let $M \to N$ be a map of $R$-modules. Let $F_\bullet \to M$ be a resolution by free $R$-modules and let $N_\bullet \to N$ be an arbitrary resolution. Then

  1. there exists a map of complexes $F_\bullet \to N_\bullet $ inducing the given map

    \[ M = \mathop{\mathrm{Coker}}(F_1 \to F_0) \to \mathop{\mathrm{Coker}}(N_1 \to N_0) = N \]
  2. two maps $\alpha , \beta : F_\bullet \to N_\bullet $ inducing the same map $M \to N$ are homotopic.

Proof. Proof of (1). Because $F_0$ is free we can find a map $F_0 \to N_0$ lifting the map $F_0 \to M \to N$. We obtain an induced map $F_1 \to F_0 \to N_0$ which ends up in the image of $N_1 \to N_0$. Since $F_1$ is free we may lift this to a map $F_1 \to N_1$. This in turn induces a map $F_2 \to F_1 \to N_1$ which maps to zero into $N_0$. Since $N_\bullet $ is exact we see that the image of this map is contained in the image of $N_2 \to N_1$. Hence we may lift to get a map $F_2 \to N_2$. Repeat.

Proof of (2). To show that $\alpha , \beta $ are homotopic it suffices to show the difference $\gamma = \alpha - \beta $ is homotopic to zero. Note that the image of $\gamma _0 : F_0 \to N_0$ is contained in the image of $N_1 \to N_0$. Hence we may lift $\gamma _0$ to a map $h_0 : F_0 \to N_1$. Consider the map $\gamma _1' = \gamma _1 - h_0 \circ d_{F, 1}$. By our choice of $h_0$ we see that the image of $\gamma _1'$ is contained in the kernel of $N_1 \to N_0$. Since $N_\bullet $ is exact we may lift $\gamma _1'$ to a map $h_1 : F_1 \to N_2$. At this point we have $\gamma _1 = h_0 \circ d_{F, 1} + d_{N, 2} \circ h_1$. Repeat. $\square$


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