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I am curious if something like the Nine Lemma ( is true in an arbitrary triangulated category. To be more explicit, suppose I have a map of cofiber sequences/distinguished triangles and I take the cofiber/mapping cone at each stage vertically (this gives a diagram like the diagram in the wikipedia link without the zeroes) then is the bottom row a cofiber sequence/distinguished triangle?

I am particularly interested in the category of spectra if that makes things easier or harder.

Also, if the result is not true in general what about when one of the maps that we end up taking the cofiber of is the identity map?

I feel like this ought to be true but I did not see anything in the two references I checked and I am not sure how to make us of verdier's/octahedral axiom.

thanks for your time.

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up vote 6 down vote accepted

Yes, a version of it is true (however, I don't think you can do it with any morphism of distinguished triangles - since the map of the cones is not unique).

The statement I know is:

Every commutative square sits inside a $4 \times 4$-diagram whose rows and colums are distinguished triangles, $8$ squares commute and one square is sign commutative.

This is called "Verdier's exercise" in the folklore and can be found in Bernstein-Beilinson-Deligne, faisceaux pervers, Proposition 1.1.11.

You can also find a proof as Lemma 2.6 in May's, The additivity of traces in tensor triangulated categories, available here.

If you want to prove it yourself, here's an outline: Start with a commutative square $ABA'B'$ and draw the diagonal $A \to B'$. Build octahedra over the two ensuing commutative triangles. Only using these octahedra, you will then be able to build a diagram of the form

$$\require{AMScd} \begin{CD} A @>>> B @>>> C @>>> A[1]\\ @VVV @VVV @VVV @VVV\\ A' @>>> B' @>>> C' @>>> A'[1]\\ @VVV @VVV\\ A'' @>>> B''\\ @VVV @VVV\\ A[1] @>>> B[1] \end{CD}$$

The morphism $C \to C'$ will be a composition of two morphisms and build yet another octahedron and complete the diagram. You'll have to rotate one triangle and that's the reason for a sign commutativity occurring in the bottom right square.

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thank you very much. and should it be a B'' in the second column and third row of your diagram? – Sean Tilson Feb 28 '11 at 17:15
@Sean: Yes, you're right. Fixed (as well as some other goofs). By the way: Amnon Neeman also discusses another version of that in his book on triangulated categories based on his notion of "good morphisms" of distinguished triangles, but I don't have the book handy, so I can't check the precise statement. – t.b. Feb 28 '11 at 17:19
I was mostly hoping to know whether or not this was indeed true and maybe get a hint. Thanks for going above and beyond! – Sean Tilson Mar 1 '11 at 3:57
@Sean: My pleasure! The main take-away from this exercise is probably the following: The morphism axiom follows from the octahedral axiom. What I was hinting at when mentioning Neeman is in fact better discussed in his article Some new axioms for triangulated categories. There he introduces a version of the mapping cone of a morphism of triangles, which allows you to get the above diagram as well. It turns out that not every morphism of triangles is such that its cone is a disinguished triangle (these are the good ones) and that essentially the octahedral axiom is equivalent to saying... – t.b. Mar 1 '11 at 10:07
...that in the morphism axiom we can find a "good" morphism, i.e., such that its mapping cone is a distinguished triangle. In a now deleted comment you were asking about Neeman's book. My (rather uninformed) opinion is that for a homotopy theorist there probably is not so much to gain from reading that book (most things are refinements of stuff you can get by techniques closer to a homotopist's usual box of tools), so you can probably spare yourself the trouble of reading it without fearing of missing much. – t.b. Mar 1 '11 at 10:09

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