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 13.4.2. Let $\mathcal{D}$ be a pre-triangulated category. For any object $W$ of $\mathcal{D}$ the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, -)$ is homological, and the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, W)$ is cohomological.

Proof. Consider a distinguished triangle $(X, Y, Z, f, g, h)$. We have already seen that $g \circ f = 0$, see Lemma 13.4.1. Suppose $a : W \to Y$ is a morphism such that $g \circ a = 0$. Then we get a commutative diagram

\[ \xymatrix{ W \ar[r]_1 \ar@{..>}[d]^ b & W \ar[r] \ar[d]^ a & 0 \ar[r] \ar[d]^0 & W[1] \ar@{..>}[d]^{b[1]} \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] } \]

Both rows are distinguished triangles (use TR1 for the top row). Hence we can fill the dotted arrow $b$ (first rotate using TR2, then apply TR3, and then rotate back). This proves the lemma. $\square$


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