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.



Proof without using Yoneda's lemma?
Yes, it is possible (compare Balmer, [Triangular Witt groups. Part I: The 12-term localization exact sequence](http://www.math.ucla.edu/~balmer/research/Pubfile/TWG-I.pdf), 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 $1-\varepsilon$.


