# 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

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllll} X & \xrightarrow{f} & Y &\rightarrow& Z\\ & & \da{\alpha} & & \\ X' & \rightarrow & Y' & \xrightarrow{k} & Z' \\ \end{array}$$ where $k\circ \alpha\circ f=0$ and horizontal rows are distinguished triangles, it is possible to find two (non unique) morphisms $\beta:X\to X$ and $\gamma: Z\to Z'$ such that

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllll} X & \xrightarrow{f} & Y &\rightarrow& Z\\ \da{\beta}& & \da{\alpha} & & \da{\gamma} \\ X' & \rightarrow & Y' & \xrightarrow{k} & Z' \\ \end{array}$$ is a morphism of distinguished triangles. I have managed to show it for an easier case, when the upper triangle is $X\to X\to 0\to X[1]$, which is distinguished by the definition of $\mathbf{D}$. Suppose that we have a diagram

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllll} X & \xrightarrow{\mathrm{id}} & X &\rightarrow& 0\\ & & \da{\alpha} & & \da{\gamma} \\ X' & \xrightarrow{j} & Y' & \xrightarrow{k} & Z' \\ \end{array}$$ such that $k\circ \alpha=0$, note that the only choice for $\gamma$ is 0. I want to show that $\alpha=j\circ \beta$ for some $\beta:X\to X'$. I can then get $\beta[1]$ from the diagram

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllll} X & \rightarrow & 0 &\rightarrow& X[1]&\xrightarrow{-\mathrm{id}}&X[1]\\ \da{\alpha} & & \da{\gamma} & & \da{\beta[1]}&&\da{\alpha[1]} \\ Y' & \xrightarrow{k} & Z' & \xrightarrow{} & X'[1]&\xrightarrow{-j[1]}& Y'[1] \\ \end{array}$$ and recover $\beta$ from $\beta[1]$. Can this construction be generalised to arbitrary distinguished triangles?

• 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. Jun 12, 2013 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. Jun 12, 2013 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. Jun 12, 2013 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.