Proof without using Yoneda's lemma?

Question

Let $\mathscr{T}$ be atriangulated category.

The third axiom for triangulated categories, namely,

if in the diagram

$$\begin{array} 0X &\stackrel{u}{\longrightarrow}&Y&\stackrel{v}{\longrightarrow}&Z&\stackrel{w}{\longrightarrow}&\Sigma X\\ \downarrow{f}&&\downarrow{g}&&\downarrow{\exists h}&&\downarrow{\Sigma f}\\ X'&\stackrel{u'}{\longrightarrow}&Y'&\stackrel{v'}{\longrightarrow}&Z'&\stackrel{w'}{\longrightarrow}&\Sigma X'\\ \end{array} $$

both rows are distinguished triangles and the left square is commutative, then there is a (not necessarily unique) map $h$ such that all the squares commute,

has the following

$\mathbf{Corollary:}$

If $f$ and $g$ are isomorphisms, so is $h$.

Now my question is:

Is there a way to prove this corollary, without using Yoneda's lemma?

Thanks for the help.

Answer 1

Yes, it is possible (compare Balmer, Triangular Witt groups. Part I: The 12-term localization exact sequence, page 5).

One easily reduces your question to the following situation: \begin{array} 0 X &\stackrel{u}{\longrightarrow} & Y & \stackrel{v}{\longrightarrow} & Z & \stackrel{w}{\longrightarrow} & \Sigma X \\ \downarrow{1} && \downarrow{1} && \downarrow{h} && \downarrow{1} \\ X&\stackrel{u}{\longrightarrow} & Y & \stackrel{v}{\longrightarrow} & Z & \stackrel{w}{\longrightarrow} & \Sigma X\\ \end{array} and the claim is that $h$ is an isomorphism.

Two observations:

1. If in an endomorphism $(f,g,h)$ of a distinguished triangle $f$ and $g$ are zero then $h^2 = 0$.

   Proof. Since $hv = 0$, we can write $h = kw$ and thus $h^2 = kwh = 0$.

2. In the above diagram, $\varepsilon = h-1$ satisfies $\varepsilon^2 = 0$ by point 1. Thus $h = 1+\varepsilon$ has the inverse $1-\varepsilon$.