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I've come across a specific problem in something that I'm working on for which I'd like to see if there's a general solution. Here's the problem stated in generality:

In a triangulated category (say, $D(R)$ the derived category of $R$-modules for a ring $R$), I have a grid of a bunch of exact triangles all pieced together that looks like this:

$$ \begin{array}{cccccc} &A&\to&B&\to&C&\xrightarrow{+1}\\ &\downarrow&&\downarrow&&\downarrow&\\ &D&\to&E&\to&F&\xrightarrow{+1}\\ &\downarrow&&\downarrow&&&\\ &G&\to&H&&&\\ {\scriptstyle+1}&\downarrow&{\scriptstyle+1}&\downarrow \end{array} $$ The maps $C\to F$ and $G\to H$ induce exact triangles from their cones. Write $C\to F\to X\xrightarrow{+1}$ and $G\to H\to Y\xrightarrow{+1}$ for these two induced triangles.

My question is this: what can be said about the relationship between $X$ and $Y$?

  1. Perhaps, being optimistic, is it the case that $X\cong Y$? (Yielding the following lovely grid where $I=X\cong Y$:)

$$ \begin{array}{cccccc} &A&\to&B&\to&C&\xrightarrow{+1}\\ &\downarrow&&\downarrow&&\downarrow&\\ &D&\to&E&\to&F&\xrightarrow{+1}\\ &\downarrow&&\downarrow&&\downarrow&\\ &G&\to&H&\to&I&\xrightarrow{+1}\\ {\scriptstyle+1}&\downarrow&{\scriptstyle+1}&\downarrow&{\scriptstyle+1}&\downarrow \end{array} $$

  1. If $X$ is not always isomorphic to $Y$, how can we "compare" the two? Is there a natural morphism between them? Which direction does it go? What properties does it have?
  2. If $X$ is not always isomorphic to $Y$, are there any categorical or homological conditions we can impose so that $X\cong Y$?
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    $\begingroup$ The scenario is the same if you interchange the horizontal and vertical directions. Hence either there is no relationship between $X$ and $Y$ or they have a symmetric relationship. My guess is that they are quasi-isomorphic, actually. $\endgroup$
    – Zhen Lin
    Aug 6, 2022 at 4:33
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    $\begingroup$ I think these should be isomorphic, if you complete to a grid in one direction, you need to show the other direction is also a triangle. I think by applying homological functors and using spectral sequences (maybe boundedness needed) you can show the other direction behaves like a triangle for any homological functor. This is barely a sketch, but it seems plausible that a proof might exist along these lines, I’ll keep thinking. $\endgroup$
    – Chris H
    Aug 6, 2022 at 6:12

1 Answer 1

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$X$ and $Y$ need not be isomorphic.

By axiom TR3 of a triangulated category, the diagram

\begin{array}{cccccc} &A&\to&B&\to&C&\xrightarrow{+1}\\ &\downarrow&&\downarrow&&&\\ &D&\to&E&\to&F&\xrightarrow{+1}\\ \end{array}

can be completed to a map of triangles

\begin{array}{cccccc} &A&\to&B&\to&C&\xrightarrow{+1}\\ &\downarrow&&\downarrow&&\downarrow&\\ &D&\to&E&\to&F&\xrightarrow{+1}\\ \end{array}

But in general the map $C\to F$ is not unique, and different choices may have non-isomorphic cones.

[For example, take the square involving $A$,$B$,$D$,$E$ to be of the form \begin{array}{ccc} &D[-1]&\to&0\\ &\downarrow&&\downarrow\\ &0&\to&D\\ \end{array} Then $C\to F$ can be chosen to be any map $D\to D$, and so $X$ can be $0$ (if we choose the identity map) or $D\oplus D[1]$ (if we choose the zero map).]

So if you fix the map $G\to H$, so that $Y$ is determined up to isomorphism, and make two different choices of the map $C\to F$ with different cones $X$, then you can't have $X\cong Y$ for both choices.

On a positive note, if you are prepared to make careful choices of the maps $C\to F$ and $G\to H$, then you do get a "lovely grid" (although extending your diagram to the bottom right, the square

\begin{array}{ccc} &I&\to&G[1]\\ &\downarrow&&\downarrow\\ &C[1]&\to&A[2]\\ \end{array}

anti-commutes rather than commuting).

This is the "triangulated $3\times3$ lemma" (or "$9$ lemma"). See for example Proposition 13.4.23 of the Stacks project.

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