# Comparing the cones of two morphisms in a triangulated/derived category

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 $$X\cong Y$$?
• 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. Aug 6, 2022 at 4:33
• 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. Aug 6, 2022 at 6:12

$$X$$ 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.