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Let $\mathscr{C}$ be an abelian category and $f:A \to B$ and $g:B \to C$ morphisms in $\mathscr{C}$, then we have the following exact sequence

$$0 \to \operatorname{Ker}(f) \to \operatorname{Ker}(gf) \to \operatorname{Ker}(g) \xrightarrow{\delta} \operatorname{Coker}(f) \to \operatorname{Coker}(gf) \to \operatorname{Coker}(g) \to 0.$$

I need to use this result which I know its true since I almost got it, but instead of looking help ending the proof. I'm wondering if someone knows a book or text where I can reference this result? This one looks pretty much like the Snake Lemma but I cannot find them in Literature. Or if someone help me see this as an application of a the Snake Lemma it would also be helpful. Thanks

Nulhomologous
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Sok
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    What's the map from $\ker(f)$ to $\ker(g)$? I don't see an obvious way to define this map—the only way you've provided to get from $A$ to $B$ is via $f$, but that sends all elements of $\ker(f)$ to zero, so how are you embedding $\ker(f)$ in $\ker(g)$? – Daniel Hast Nov 21 '20 at 16:59
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    You must be missing some hypotheses. Otherwise take $B=C=0$ and $f=g=0$. Then the sequence would have to look like $0\rightarrow A\rightarrow 0\rightarrow...$, forcing $A$ to be zero. – Thorgott Nov 21 '20 at 17:01
  • Sorry! It was a typos mistake, already edited my mistake. I wanted to say $0\to Ker(f) \to Ker(gf) \to Ker(f) \to....$ – Sok Nov 21 '20 at 17:08
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    I guess you wanted to say $$ 0 \to \operatorname{Ker}(f) \to \operatorname{Ker}(gf) \to \operatorname{Ker}(g) \to \dots $$ – Nulhomologous Nov 21 '20 at 17:55
  • See also https://math.stackexchange.com/questions/436112/is-this-exact-sequence-a-special-case-of-the-snake-lemma – Robert Bruner Dec 30 '20 at 20:21
  • If you just wish for a reference, this result is stated in Milne's CFT notes as lemma A.2 (called there the "kernel-cokernel lemma"). – Wojowu Aug 24 '21 at 18:17

2 Answers2

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It is a consequence of the snake lemma applied to $$\array{ \operatorname{Ker}(f) &\to & A &\stackrel{f}{\to} & B &\to& \operatorname{Coker}(f) &\to & 0 \\ && \downarrow^{gf} && \downarrow^{g} && \downarrow \\ 0 &\to& C &\to & C &\to& 0 } $$ and you get $$ 0 \to \operatorname{Ker}(f) \to \operatorname{Ker}(gf) \to \operatorname{Ker}(g) \to \operatorname{Coker}(f) \to \operatorname{Coker}(gf) \to \operatorname{Coker}(g) \to 0$$

Nulhomologous
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  • Thanks! But wondering how you connect $Ker(f)$ with $Ker(gf)$? Same with $Coker(f)$ with $Coker(gf)$? @Nullhomologous – Sok Nov 21 '20 at 18:03
  • They are just the natural maps (in terms of elements): an element in $Ker(f)$ is inside $Ker(gf)$, as if $f(x)=0$, then $g(f(x))=g(0)=0$. The map from $Coker(f)$ to $Coker(gf)$ are the connecting maps: take a lifting to B, apply $g$, go down to $Coker(gf)$. – Nulhomologous Nov 21 '20 at 18:07
  • Already understand the morphism $Ker(f)$ into $Ker(gf)$ but stil dont see how the conencting map from $Ker(g)$ to $Coker (f)$? Im thinking in a diagram this way: https://en.wikipedia.org/wiki/Snake_lemma. So in my head im trying to get the connecting map from $Ker(Coker(f) \to 0)$ to Coker(gf) @Nullhomologous – Sok Nov 21 '20 at 18:24
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    The map from $Ker(g) $ to $Coker(f) $ in the presentation I put is just the composition $Ker(g)\to B \to Coker(f)$. – Nulhomologous Nov 23 '20 at 08:26
  • Thanks! I think I already got your idea! Nulhomologous – Sok Nov 23 '20 at 17:18
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Try applying the snake lemma to the diagram $$ \require{AMScd} \begin{CD} 0 @>>> A @>{\begin{bmatrix} id \\ f \end{bmatrix}}>> A \oplus B @>{\begin{bmatrix} -f & id \end{bmatrix}}>> B @>>> 0 \\ @. @VVfV @VV{\begin{bmatrix} 0 & id \\ gf & 0 \end{bmatrix}}V @VV{-g}V \\ 0 @>>> B @>{\begin{bmatrix} id \\ g \end{bmatrix}}>> B \oplus C @>{\begin{bmatrix} -g & id \end{bmatrix}}>> C @>>> 0. \end{CD} $$

Daniel Schepler
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