Cones of a morphism of distinguished triangles In a derived category (actually, I am interested in the derived category of abelian groups, if that helps), suppose we are given a morphism of distinguished triangles
$$\require{AMScd}
\begin{CD}
X @>u>> Y @>v>> Z @>w>> X [1] \\
@VVV @VVV @VVV @VVV \\
X' @>>> Y' @>>> Z' @>>> X' [1]
\end{CD}$$
We may pick the cone of each vertical arrow and obtain a diagram
$$\require{AMScd}
\begin{CD}
X @>u>> Y @>v>> Z @>w>> X [1] \\
@VVV @VVV @VVV @VVV \\
X' @>>> Y' @>>> Z' @>>> X' [1] \\
@VVV @VVV @VVV @VVV \\
X'' && Y'' && Z'' && X'' [1] \\
@VVV @VVV @VVV @VVV \\
X [1] && Y [1] && Z [1] && X [2]
\end{CD}$$
And then using the axiom (TR3), let's complete this to a commutative diagram
$$\require{AMScd}
\begin{CD}
X @>u>> Y @>v>> Z @>w>> X [1] \\
@VVV @VVV @VVV @VVV \\
X' @>>> Y' @>>> Z' @>>> X' [1] \\
@VVV @VVV @VVV @VVV \\
X'' @>>> Y'' @>>> Z'' @>>> X'' [1] \\
@VVV @VVV @VVV @VVV \\
X [1] @>u [1]>> Y [1] @>v [1]>> Z [1] @>w [1]>> X [2]
\end{CD}$$
Here the columns are distinguished triangles and the first two rows are distinguished triangles; the bottom row is not a distinguished triangle, because the signs of its morphisms are off, but it still induces a long exact sequence in cohomology; finally, for the obtained row $X'' \to Y'' \to Z'' \to X'' [1]$ there is no reason to be distinguished at all.
But still, can we say something about $X'' \to Y'' \to Z'' \to X'' [1]$?
1) Do its arrows compose to $0$?
2) Does it give a long exact sequence in cohomology?
P.S. Just in case, I am aware that the nine-lemma in triangulated categories does not work in this naive fashion (see Nine lemma in Triangulated categories).
Thank you.
 A: In the derived category of vector spaces over a field $k$, take the diagram
$$\require{AMScd}
\begin{CD}
k[-1] @>>> 0 @>>> k @>\text{id}>> k \\
@VVV @VVV @VV0V @VVV \\
0 @>>> k @>\text{id}>> k @>>> 0\\
@VVV @V\text{id}VV @V\pmatrix{1&0}VV @VVV \\
k @. k @. k\oplus k[1] @. k[1] \\
@V\text{id}VV @VVV @V\pmatrix{0\\1}VV @V\text{id}VV \\
k @.0 @. k[1] @. k[1]
\end{CD}$$
where the rows and columns are distinguished triangles.
This can be completed to a commutative diagram
$$\begin{CD}
k[-1] @>>> 0 @>>> k @>\text{id}>> k \\
@VVV @VVV @VV0V @VVV \\
0 @>>> k @>\text{id}>> k @>>> 0\\
@VVV @V\text{id}VV @V\pmatrix{1&0}VV @VVV \\
k @>\text{id}>> k @>\pmatrix{1&0}>> k\oplus k[1] @>\pmatrix{0\\1}>> k[1] \\
@V\text{id}VV @VVV @V\pmatrix{0\\1}VV @V\text{id}VV \\
k @>>> 0 @>>> k[1] @>\text{id}>> k[1]
\end{CD}$$
where the first two maps in the third row do not compose to zero, even on cohomology.
Here's an example where the maps do compose to zero, but the third row does not have a long exact sequence of cohomology:
$$\begin{CD}
k[-1] @>\text{id}>> k[-1] @>>> 0 @>>> k \\
@V\text{id}VV @VVV @VVV @V\text{id}VV \\
k[-1] @>>> 0 @>>> k @>\text{id}>> k\\
@VVV @VVV @V\text{id}VV @VVV \\
0 @>>> k @>0>> k @>>> 0 \\
@VVV @V\text{id}VV @VVV @VVV \\
k @>\text{id}>> k @>>> 0 @>>> k[1]
\end{CD}$$
The columns, and all the rows except the third one, are distinguished triangles. But taking cohomology of the third row gives
$$\dots\to0\to k\stackrel{0}{\to}k\to0\to\dots,$$
which is not exact.
