$0—>A—>B—>C—>0$ is a short exact sequence of differential complexes. I wonder why for any element $c$ belonging to $C$, it is true that $dc=0$.
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Does $d$ denote the differential $C\xrightarrow{d} 0$? – Zest Nov 09 '20 at 07:16
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@Zest No. $d$ is differential operator define on differential complex C. – Midas Hu Nov 09 '20 at 07:21
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In the context of differential complexes, the notation d typically represents the differential operator, which maps an element of one complex to the next. The property dc=0 for all c in C is a consequence of the exactness of the sequence at C . A short exact sequence of complexes 0→A→B→C→0 is exact at C if the image of the differential map from B to C is equal to the kernel of the differential map from C to 0 . But the kernel of any map to the zero object 0 is the entire domain, so in this case, it’s all of C . Therefore, the image of the differential map from B to C must also be all of C . However, the only element that maps to 0 under the differential operator is 0 itself (since we’re in a complex, where the composition of two consecutive differentials is zero: d2=0 ). Therefore, for all c in C , we must have dc=0
