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\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 10.108.1 (Schanuel's lemma). Let $R$ be a ring. Let $M$ be an $R$-module. Suppose that

\[ 0 \to K \xrightarrow {c_1} P_1 \xrightarrow {p_1} M \to 0 \quad \text{and}\quad 0 \to L \xrightarrow {c_2} P_2 \xrightarrow {p_2} M \to 0 \]

are two short exact sequences, with $P_ i$ projective. Then $K \oplus P_2 \cong L \oplus P_1$. More precisely, there exist a commutative diagram

\[ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} \ar[d] & P_1 \oplus P_2 \ar[r]_{(0, p_2)} \ar[d] & M \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(p_1, 0)} & M \ar[r] & 0 } \]

whose vertical arrows are isomorphisms.

Proof. Consider the module $N$ defined by the short exact sequence $0 \to N \to P_1 \oplus P_2 \to M \to 0$, where the last map is the sum of the two maps $P_ i \to M$. It is easy to see that the projection $N \to P_1$ is surjective with kernel $L$, and that $N \to P_2$ is surjective with kernel $K$. Since $P_ i$ are projective we have $N \cong K \oplus P_2 \cong L \oplus P_1$. This proves the first statement.

To prove the second statement (and to reprove the first), choose $a : P_1 \to P_2$ and $b : P_2 \to P_1$ such that $p_1 = p_2 \circ a$ and $p_2 = p_1 \circ b$. This is possible because $P_1$ and $P_2$ are projective. Then we get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} & P_1 \oplus P_2 \ar[r]_{(0, p_2)} & M \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[d] \ar[u] & P_1 \oplus P_2 \ar[r]_{(p_1, p_2)} \ar[d]_ T \ar[u]^ S & M \ar[r] \ar@{=}[d] \ar@{=}[u] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(p_1, 0)} & M \ar[r] & 0 } \]

with $T$ and $S$ given by the matrices

\[ S = \left( \begin{matrix} \text{id} & 0 \\ a & \text{id} \end{matrix} \right) \quad \text{and}\quad T = \left( \begin{matrix} \text{id} & b \\ 0 & \text{id} \end{matrix} \right) \]

Then $S$, $T$ and the maps $N \to P_1 \oplus L$ and $N \to K \oplus P_2$ are isomorphisms as desired. $\square$

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