2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Avitus Jun 12 '13 at 20:17
  • $\begingroup$ 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. $\endgroup$ – Jimmy R Jun 12 '13 at 20:38
  • $\begingroup$ 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. $\endgroup$ – Avitus Jun 12 '13 at 20:46
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks! I've just arrived at a similar solution myself. $\endgroup$ – Jimmy R Jun 12 '13 at 21:21
  • $\begingroup$ you are welcome! $\endgroup$ – Avitus Jun 13 '13 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.