# Embedding into a morphism of distinguished triangles

Everything in this question happens in a triangulated category $\mathbf{D}$. I am trying to prove that in a diagram like this





• The existence of (in general not unique) $\beta$ s.t. $\alpha=j\circ \beta$ in the case $X\rightarrow X\rightarrow 0$ above is a consequence of axiom T3 of triangulated categories. – Avitus Jun 12 '13 at 20:17
• I agree, $\beta[1]$ is made by completion of the diagram using T3 and then $\beta$ is obtained using T2, which states that a triangle can be started anywhere. – Jimmy R Jun 12 '13 at 20:38
• I agree: I checked in Neeman's book and also there the morphism whose existence is axiomatized is the 3rd. I tried to construct a canonical triangle using $\alpha\circ f$ and the axiom T4 but I arrived at a sequence $Y\stackrel{\alpha}{\rightarrow} Y'\stackrel{1}{\rightarrow}Y'$ which cannot be a distinguished triangle. – Avitus Jun 12 '13 at 20:46

Use cohomological functors: $D$ is triangulated, so $Hom(X,)$ is a cohomological functor for anz object $X$ in $D$. In particular, the short sequence
$Hom(X,X')\rightarrow Hom(X,Y')\rightarrow Hom(X,Z')$ (*)
is exact,with $X'\stackrel{q}{\rightarrow} Y'\rightarrow Z'$ distinguished in $D$.
As $\alpha\circ f$ belongs to the kernel of $Hom(X,Y')\rightarrow Hom(X,Z')$ by hypothesis, then there exists a $r\in Hom(X,X')$ s.t. $q\circ r= \alpha\circ f$ by exactness of (*).
The last morphism between the distinguished triangles $X\rightarrow Y\rightarrow Z$ and $X'\stackrel{q}{\rightarrow} Y'\rightarrow Z'$ can be found using the axiom T3.