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About the Exercise 1.2.4 enter image description here

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?

闫嘉琦
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  • Just a moment, is $\mathcal A$ an arbitrary Abelian category? If so, the definition of exactness is not simply "\ker(g)=\im(f)$, the isomorphism between them must be the one that is naturally induced. – Just dropped in Apr 04 '20 at 14:21
  • @JustDroppedIn Yes, according to the previous pages, $\mathcal{A}$ is an arbitrary abelian category, and $\mathbf{Ch}=\mathbf{Ch(\mathcal{A})}$ is the category of chain complexes in $\mathcal{A}$. – 闫嘉琦 Apr 04 '20 at 14:23
  • You're meant to prove that $\text{ker}(f) = {\text{ker}(f_n) }$ because that is not a trivial statement. – Noel Lundström Apr 04 '20 at 16:07
  • @NoelLundström Yes, I have done this. Use the commutative diagram and the universal property to induce a map $d_{}:\ker{f_n}\to\ker{f_{n-1}}$ by $d:A_n\to A_{n-1}$. And this also shows that $d_{}$ and $f_n$ commute. And we can show that $d_{*}^2=0$ and $\ker{f}={\ker{f}_n}$ also has universal property. Then $ker(f)$ is a chain complex and is the kernel of $A.\to B.$. – 闫嘉琦 Apr 04 '20 at 16:14
  • @NoelLundström I think this is enough for $\ker(f)={\ker(f_n)}$. – 闫嘉琦 Apr 04 '20 at 16:17
  • Yes that is correct. And yes with that fact for $\text{ker}(f)$ and a similar fact for $\text{coker}(f)$ your solution in your post is complete! – Noel Lundström Apr 04 '20 at 16:20

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