6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Steve
    Jul 20, 2012 at 20:31

2 Answers 2

2
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Sasha
    Jul 22, 2012 at 4:20
  • $\begingroup$ @Yosemite Sam: this is clearly not enough. How could you recover the isomorphism class of $E$ from its cohomology ? $\endgroup$
    – Niels
    Jul 21, 2016 at 20:40
  • $\begingroup$ @Niels this was a long time ago so I might be missing something: if E is such that $H^j(E)=0$ for all j not equal to some fixed i, then E = H^i(E)[-i]. Isn't that enough? $\endgroup$ Aug 19, 2016 at 16:33
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.