How to compute connecting homomorphism?

Question

Let $A_.$ be the complex $\dots\to\mathbb{Z}^2\to\mathbb{Z}^2\to\mathbb{Z}^2\to\dots$ with morphisms $d_{A,n}((a,b))=(b,0)$.

Let $B_.$ be the complex $\dots\to\mathbb{Z}^3\to\mathbb{Z}^3\to\mathbb{Z}^3\to\dots$ with morphisms $d_{B,n}((a,b,c))=(b,0,0)$.

Let $C_.$ be the complex $\dots\to\mathbb{F}_2^2\to\mathbb{F}_2^2\to\mathbb{F}_2^2\to\dots$ with morphisms $d_{C,n}((a,b))=(b,0)$.

The chain maps $f:A_.\to B_.$ and $g:B_.\to C_.$ defined by $f(a,b)=(2a,2b,0)$ and $g(a,b,c)=(a,b)$ produces an exact sequence of chain complexes $0\to A_.\to B_. \to C.\to 0$. What is the connecting homomorphism $\partial:H_n(C_.)\to H_{n-1}(A_.)$? How is it computed?

Answer 1

There is a canonical definition for the connecting homo orphism as follows: It is \partial [c]=[a] where the equivalence class of a is the obtained as follows: pick an element in b mapping to c then map b via d_B,n to another element d_B,n(b) in B, which maps to zero after applying g. But this means that there is an a in A mapping to d_B,n(b) via. f i.e.f(a)=d_B,n(b). The equivalence class [a] is independent of the choices made.

You are however in trouble as your candidate for complex C is not a complex ( since the image is not contained in the kernel)