L = longExactSequence(g, f)
Every short exact sequence of complexes
$\phantom{WWWW} 0 \leftarrow C \xleftarrow{g} B \xleftarrow{f} A \leftarrow 0 $
gives rise to a long exact sequence $L$ in homology having the form
$\phantom{WWWW} \dotsb \leftarrow H_{i-1}(C) \xleftarrow{H_{i-1}(g)} H_{i-1}(B) \xleftarrow{H_{i-1}(f)} H_{i-1}(A) \xleftarrow{\delta_i} H_{i}(C) \xleftarrow{H_{i}(g)} H_{i}(B) \xleftarrow{H_{i}(f)} H_{i}(A) \leftarrow \dotsb. $
This method returns the complex $L$ such that, for all integers $i$, we have $L_{3i} = H_i(C)$, $L_{3i+1} = H_i(B)$, and $L_{3i+2} = H_i(A)$. The differentials $\operatorname{dd}^L_{3i}$ are the connecting homomorphisms $\delta_i \colon H_i(C) \to H_{i-1}(A)$. Moreover, we have $\operatorname{dd}^L_{3i+1} = H_{i}(g)$ and $\operatorname{dd}^L_{3i+2} = H_{i}(f)$.
For example, consider a free resolution $F$ of $S/I$. Applying the Hom functor $\operatorname{Hom}(F, -)$ to a short exact sequence of modules
$\phantom{WWWW} 0 \leftarrow S/h \leftarrow S \xleftarrow{h} S(- \deg h) \leftarrow 0 $
gives rise to a short exact sequence of complexes. The corresponding long exact sequence $L$ in homology has the form
$\phantom{WWWW} \dotsb \leftarrow \operatorname{Ext}^{d+1}(S/I, S(-\deg h)) \xleftarrow{\delta} \operatorname{Ext}^d(S/I, S/h) \leftarrow \operatorname{Ext}^d(S/I, S) \leftarrow \operatorname{Ext}^d(S/I, S(-\deg h)) \leftarrow \dotsb. $
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We verify that the indexing on $L$ in this example matches the description above.
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