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$\renewcommand{\im}{\mathop{\rm im}}\DeclareMathOperator{\coker}{coker}$Let $A\xrightarrow{f}B\xrightarrow{g}C$ be a complex in an abelian category, I.e. $gf=0$. Let $H:=\coker(\im(f)\to \ker(g)).$

Then $H\xrightarrow{i}\coker(f)\xrightarrow{p}\im(g)$ is exact.

Could you give me a proof of it?

I can say that $i$ is a monomorphism and $p$ is epi. I know that $\im(f)=\ker(\coker(f))=\coker(\ker(f))$ and that $\im(f)\to \ker(g)$ is monomorphic.

But I want $\im(i)=\ker(p).$ I cannot prove even that $pi=0.$

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3  
See Kashiwara-Schapira, (8.3.4), page 178. This thread might be interesting (although it doesn't contain an answer to your specific question). –  commenter Oct 27 '12 at 13:22
    
Do you accept usage of the nine lemma here? Your desired sequence is obtained by taking $\im(f)$-quotient of the exact sequence $0\to \ker(g)\to B\to \im(g)\to 0$ which may be the bottom row of an $3\times 3$ square with exact cols and two exact top rows. –  Ben A. Oct 27 '12 at 16:06
    
@commenter Thank you for letting know me the book. –  Tom Oct 28 '12 at 6:00

2 Answers 2

up vote 2 down vote accepted

Let us denote the objects by capital and corresponding arrows by small letters: ${\rm coim} f:A\to {\rm Coim} f$. Sorry, I write from left to right.

First of all, as in any Abelian category, $f={\rm coim} f\cdot {\rm im}f$, so $$fg=0 \iff {\rm coim}f\, {\rm im}f\cdot{\rm coim}g\,{\rm im}g=0 \iff {\rm im}f\cdot {\rm coim g}=0,$$ as ${\rm coim} f$ is epi and ${\rm im}g$ is mono, moreover, we also have ${\rm ker}g={\rm ker}({\rm coim}g)$ and ${\rm coker}f={\rm coker}({\rm im}f)$. Hence wlog. we can assume that $f$ is mono ($f={\rm im}f$) and $g$ is epi ($g={\rm coim} g$), and this way $A={\rm Im}f$ and $C={\rm Coim}g$ is also assumed.

We can then draw a (bit simplified) diagram with $$ A\underbrace{\overset{j}\to {\rm Ker}g \to B}_f \overbrace{\to{\rm Coker}f \underset{p}\to C}^g$$ where $H={\rm Coker}j$. As $j\cdot{\rm ker}g\cdot{\rm coker}f=f\cdot{\rm coker}f=0$, we get the $i:H\to{\rm Coker}f$. Now $${\rm coker}j\cdot i\, p= {\rm ker}g\cdot {\rm coker}f\cdot p= {\rm ker}g\cdot g=0$$ hence, $ip=0$, because ${\rm coker}j$ is epi.

For the exactness, it is enough to prove that $p={\rm coker}i$. For this: if $iv=0$ for any arrow $v$, we have $0={\rm coker} j\cdot iv={\rm ker} g\cdot{\rm coker}f\cdot v$. But, as $g$ is epi, $g={\rm coim}g={\rm coker}({\rm ker}g)$, this goes through $g$: we get a $t$ such that $gt={\rm coker}f\cdot v$. Then, using that ${\rm coker}f$ is epi, we arrive to $pt=v$.

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Thank you very much for writing the detailed answer. I understood all steps without any trouble! –  Tom Oct 28 '12 at 3:31
    
Only for our statement, the first argument showing $f$ mono $g$ epi does not contribute to shorten the total length of the proof, doesn't it? However the argument was first for me and very informative, thank you so much. –  Tom Oct 28 '12 at 5:57

$\renewcommand{\im}{\mathop{\rm im}}\DeclareMathOperator{\coker}{coker}$You can use Freyd-Mitchell Embedding theorem to reduce to the case of modules. Note that $H = \ker(g)/\im(f)$, so you have $$\ker(g)/\im(f) \xrightarrow{i}\coker(f) = B/\im(f) \xrightarrow{p} \im(g).$$ Now, clearly $pi=0$, so then we have that $\im i \subset \ker p$. For the opposite inclusion suppose that $x \in B /\im(f)$ is in $\ker p$. Take a representative element of the coset in B, and argue that it must lie in $\ker(g)$. Argue that you always can lift a coset x in $B/\im(f)$ that lies in the kernel of p to an element of $\ker(g)/\im(f)$.

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Thank you. I know the proof for modules. I hope a direct proof without their big theorem. –  Tom Oct 27 '12 at 13:03
    
Happy to help! Good luck. –  Dedalus Oct 27 '12 at 13:03
    
Do you have a direct proof? –  Tom Oct 27 '12 at 13:04

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