3
$\begingroup$

If $\mathcal{H}$ is the heart of a bounded t-structure in a triangulated category $\mathcal{T}$, then for every object $E$ in $\mathcal{T}$ there exists a finite sequence of integers $k_1>k_2>\dots >k_n$ and a collection of triangles in $\mathcal{T}$ $$ E_{i-1}\stackrel{\phi_i}{\to}E_i\to A_i \to E_{i-1}[1] $$ for $i=0,\dots,n$, with $E_0=0$, $E_n=E$ and $A_i\in \mathcal{H}[k_i]$ for any $i=1,\dots,n$. This can be seen as a Postnikov tower for the initial morphism $0 \to E$, so by analogy with what happens in algebraic topology one is led to wonder whether more generally one has Postnikov towers for arbitrary morphisms in a bounded t-structure, i.e., the same as above, but this time with $E_0=X$, $E_n=Y$ and $\phi_n\circ\phi_{n-1}\circ\cdots\circ \phi_1$ equal to some given morphism $f:X\to Y$.

I have so far been unable to rigorously prove (or disprove) such a statement nor to locate it in the literature. As usual any suggestion in either direction will be appreciated.

$\endgroup$

1 Answer 1

5
$\begingroup$

Let $F$ be the fiber of $f:X \to Y$ (i.e. the cone shifted by $[-1]$) so that we have a distinguished triangle $$ F \to X \to Y. $$ Let $0 = F_0 \to F_1 \to \dots \to F_n = F$ be its Postnikov tower. Let $E_i$ be the cone of the map $F_i \to X$ (the composition of $F_i \to F$ and $F \to X$). Then $E_0 = Cone(0 \to X) = X$ and $E_n = Cone(F \to X) \cong Y$. On the other hand, by octahedron axiom $$ Cone(E_{i-1} \to E_i) = Cone (F_{i-1} \to F_i)[1], $$ so I think this is what you want.

$\endgroup$
1
  • $\begingroup$ Thanks! Now that I see it, I see I had proved this a couple of months ago and then I had completely forgotten the reasoning (the Kubla Khan effect :) ). Thanks a lot! $\endgroup$ May 11, 2014 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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