# Question about a proof of mapping cones and the octohedral axiom for a triangulated category

I am reading a proof in a paper of Neeman's, Some New Axioms for Triangulated Categories. It can be found here. Let $\mathfrak{T}$ be a triangulated category. There he defines a subcategory $CT(\mathfrak{T})$ of so-called candidate triangles, consisting of sequences, $$X \stackrel{u}{\longrightarrow} Y \stackrel{v}{\longrightarrow} Z \stackrel{w}{\longrightarrow} \Sigma X$$ with $v \circ u$, $w \circ v$, and $\Sigma u \circ w$ all being the zero morphism. Morphisms in this subcategory are defined in exactly the way that morphisms of triangles are defined: As $\Sigma$-periodic morphisms making the resulting diagram commute.

A homotopy of two such morphisms is defined in the natural way. That is, they respect this $\Sigma$-periodicity.

Finally, he define a full subcategory of $CT(\mathfrak{T})$, denoted by $T(\mathfrak{T})$ consisting of so-called true triangles, which have as objects candidate triangles which are also distinguished triangles provided by the triangulated structure on $\mathfrak{T}$.

The following is Lemma 1.4 in the paper linked above.

Let $X^{\bullet}$ and $Y^{\bullet}$ be objects in $T(\mathfrak{T})$. Then the cone on the zero map $C(X^{\bullet} \stackrel{0}{\rightarrow} Y^{\bullet})$ is in $T(\mathfrak{T})$.

To prove this, he considers two objects in $T(\mathfrak{T})$, $$X \stackrel{u}{\longrightarrow} Y \stackrel{v}{\longrightarrow} Z \stackrel{w}{\longrightarrow} \Sigma X$$ and $$X' \stackrel{u'}{\longrightarrow} Y' \stackrel{v'}{\longrightarrow} Z' \stackrel{w'}{\longrightarrow} \Sigma X'.$$ He then claims that we need to show that the diagram $$X' \oplus X' \longrightarrow Y' \oplus Y' \longrightarrow Z' \oplus Z' \longrightarrow \Sigma X \oplus \Sigma X'$$ is a triangle.

I don't see how this proves the claim at all. What does the above sequence have anything to do with the mapping cone in question? The cone is the diagram $$X' \oplus Y \longrightarrow Y' \oplus Z \longrightarrow Z' \oplus \Sigma X \longrightarrow \Sigma X \oplus \Sigma Y.$$ Can anyone explain this proof to me?

Recall that, in Neeman's notation, $\tilde{\Sigma}$ is the operation on candidate triangles that takes $$X\stackrel{u}{\longrightarrow}Y\stackrel{v}{\longrightarrow}Z\stackrel{w}{\longrightarrow}\Sigma X$$ to $$Y\stackrel{-v}{\longrightarrow}Z\stackrel{-w}{\longrightarrow}\Sigma X\stackrel{-\Sigma u}{\longrightarrow}\Sigma Y,$$ and that if $X^\bullet$ is a triangle, then $\tilde{\Sigma}X^\bullet$ is also a triangle.
The cone of the zero map $X^\bullet\to Y^\bullet$ between two triangles is precisely $\tilde{\Sigma}X^\bullet\oplus Y^\bullet$, so to prove that the cone of a zero map between triangles is a triangle, it suffices to prove that the direct sum of two triangles is a triangle.