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.