Composition of Homomorphisms In An Exact Sequence
Theorem
Given an exact sequence
\[ M_0 \xrightarrow{f_1} M_1 \xrightarrow{f_2} M_2 \xrightarrow{f_3} M_3 \xrightarrow{f_4} \dots \xrightarrow{f_n} M_n\]
\(f_{i + 1} \circ f_i = 0\) for any \(i \in \{1, \dots, n - 1\}\).
Proof
This follows directly from the definition of an exact sequence. For any \(x \in M_{i - 1}\), \(f_i(x) \in \mathrm{im}(f_i) = \ker(f_{i + 1})\) and therefore \(f_{i + 1}(f(x)) = 0\) by definition of the kernel.