Nine lemma in Triangulated categories I am curious if something like the Nine Lemma (http://en.wikipedia.org/wiki/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.
 A: 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.
