When proving Lemma 1.4.4 in his book on triangulated categories, Neeman asserts that a certain mapping cone splits into a direct sum of three candidate triangles. I'm unable to see why this is so. Does anyone see why this holds?

Have you tried using Lemma 1.2.4? Seems like two applications of it would do the trick.
– t3sujiNov 19 '10 at 15:41

It is not immediately clear to me how Lemma 1.2.4 can by applied twice to get the desired splitting of the mapping cone. Nevertheless, I would try massaging Lemma 1.2.4 to see if two doses of a modified form will produce the desired splitting.
– Beren SandersNov 19 '10 at 21:03

1

It is not immediately clear because it is not completely spelled out: basically, you need to know something about the triangle you get after the first application. At the end of the proof of Lemma 1.2.4 he says `... the maps [in the direct summand triangle] are all very explicitly computable'. The suggested way of computation (as a kernel) implies that the arrow $Z\to\Sigma X$ in the result is the restriction of the map $A\oplus Z\to\Sigma X$ in the original triangle. (I think.) This is enough (in Lemma 1.4.4, split off $Y$ first, and then $Y'$).
– t3sujiNov 19 '10 at 22:22

2

By the way, it's both counter-productive, and somewhat rude to ask about a theorem by number, without providing the statement. Maybe someone who doesn't have Neeman's book knows how to prove the theorem none the less.
– Ben Webster♦Nov 20 '10 at 13:07

4

@Ben: I agree with you in principle (I just happened to have Neeman's book on my desk), but in this case, it is really hard to provide enough details. The question concerns one step in a proof of a (somewhat technical) lemma, the proof is about one page, and might also depend on some terminology that is specific to the book. Short of retyping a page out of the book, I do not see what else the OP could have done.
– t3sujiNov 20 '10 at 15:58