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Let $\mathcal{A}$ be an abelian category and let $$ 0 \rightarrow E \rightarrow F \rightarrow G \rightarrow 0 $$ be a short exact sequence. Then in $D(\mathcal{A})$, the derived category of $\mathcal{A}$ we have a distinguished triangle $$ E \rightarrow F \rightarrow G \rightarrow E[1]. $$

Moreover, the above short exact sequence determines an element of Ext$^1(G,E)$ and therefore a map $G[-1] \rightarrow E$ in $D(\mathcal{A})$. The axiom TR2 tells us that the cone of this map is exactly $F$.

I would like to understand if one can generalise this picture.

Consider a morphism $f: A \rightarrow B$ in $\mathcal{A}$ such that both Ker$(f)$ and Coker$(f)$ are non-zero. $f$ can be also thought of as a morphism in $D(\mathcal{A})$. What is its cone then? Do we have a distinguished triangle $$ A \rightarrow B \rightarrow \textrm{Coker}(f) \oplus \textrm{Ker}(f)[1] \rightarrow A[1]? $$

On the other hand, we have the following exact sequence in $\mathcal{A}$ $$ 0 \rightarrow \textrm{Ker}(f) \rightarrow A \rightarrow B \rightarrow \textrm{Coker}(f) \rightarrow 0. $$ It corresponds to an element in $\textrm{Ext}^2(\textrm{Coker}(f), \textrm{Ker}(f))$ and hence to a map $\textrm{Coker}(f)[-2] \rightarrow \textrm{Ker}(f)$ in $D(\mathcal{A})$. Can one write down the cone of this map using objects $A$ and $B$? My only guess is that it would be $A \oplus B[-1]$, but I don't know whether it is correct or how to prove it.

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The answer to your first question is yes if A is a hereditary category (meaning the higher Ext groups vanish), but it is not true in general. –  Steve Jul 20 '12 at 20:31

2 Answers 2

up vote 2 down vote accepted

I'll expand on Steve's comment, but I'm sure you must know this stuff already.

if $f: A \to B$ you take the cone $D$ and cohomology long exact sequence tells you that the cohomology of $D$ is $K = \text{ker } f$ in degree minus one and $C = \text{coker } f$ in degree zero. Explicitly a cone is given by the mapping cone construction for the category of complexes. The complex is given by $A[1] \oplus B$ but with the differential twisted by $f$ (actually, as these are complexes concentrated in a single degree, the differential is $f$). The abelian category being hereditary should be equivalent to saying that mapping cones split, in the sense that they can be rewritten as a direct sum of $K$ and $C$ as you wrote.

As for your second question, maybe this helps. Consider the inclusion $K \to [A \to B]$ where I'm thinking of the second guy as the complex where $A$ sits in degree zero, $B$ in degree one and the differential is $f$ (in other words it's the mapping cone of $f$ shifted by minus one). Take the cone $E$. The merciful lord of cohomology long exact sequences then tells us

$$ 0 \to K \to K \to H^0(E) \to 0 \to C \to H^1(E) \to 0 $$ and, as the first map is an isomorphism, $E$ is actually $C[-1]$. Is this enough?

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So, summarizing there are two distinguished triangles $$ A \to B \to D $$ (the definition of $D$), and $$ K[1] \to D \to C $$ (the canonical filtration on $D$). The second extends to give a map $C\to K[2]$ which is the corresponding extension class. –  Sasha Jul 22 '12 at 4:20

Not a real answer, but hopefully a relevant comment:

There is s very close analogy between the homotopy category of spaces and chain complexes. You can find such presentation in Peter May's notes on chain complexes I belive. Under this correspondence cones correspond to cones. Now there is something called a Dold-Puppe sequence, you can find it's description in Hatcher's book. In this sequence every map is a cone of a previous one and these sequences correspond exactly to distinguished triangles.

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