About the Exercise 1.2.4
Suppose that $A.\xrightarrow{f} B.\xrightarrow{g} C.$
I know that $\ker{f}=\{\ker{f_n}\}$, $\mathrm{im}{f}=\{\mathrm{im} f_n\}$ are chain complexes of the abelian category $\mathcal{A}$, where $\ker\{f_n\}\xrightarrow{\ker{f_n}} A_n$ is the kernel of $A_n\xrightarrow{f_n} B_n$. $\mathrm{im} f_n=\ker(\mathrm{coker}{f_n})$ is defined similarly.
Then the sequence in $\mathbf{Ch}$ is exact at $B.$ $\iff$ $\ker{g}=\mathrm{im}{f}$
$\iff$ $\ker{f}_n=\mathrm{im}{f}_n$ for each $n\in\mathbb{Z}$
$\iff$ the sequence in $\mathcal{A}$ is exact at $B_n$ for each $n\in\mathbb{Z}$.
My question is, is this enough to prove this exercise? And if not, could you please help me to fill the gaps?