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 10.87.5. Let $f: M \to N$ and $g: M \to M'$ be maps of $R$-modules. Suppose $\mathop{\mathrm{Coker}}(f)$ is of finite presentation. Then $g$ dominates $f$ if and only if $g$ factors through $f$, i.e. there exists a module map $h: N \to M'$ such that $g = h \circ f$.

Proof. Consider the pushout of $f$ and $g$ as in the statement of Lemma 10.87.4. From the construction of the pushout it follows that $\mathop{\mathrm{Coker}}(f') = \mathop{\mathrm{Coker}}(f)$, so $\mathop{\mathrm{Coker}}(f')$ is of finite presentation. Then by Lemma 10.81.4, $f'$ is universally injective if and only if

\[ 0 \to M' \xrightarrow {f'} N' \to \mathop{\mathrm{Coker}}(f') \to 0 \]

splits. This is the case if and only if there is a map $h' : N' \to M'$ such that $h' \circ f' = \text{id}_{M'}$. From the universal property of the pushout, the existence of such an $h'$ is equivalent to $g$ factoring through $f$. $\square$


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