7
$\begingroup$

Let $$ \begin{array}{rcl} A&\to& B\\ \downarrow & &\downarrow\\ A'&\to& B' \end{array} $$ be a commutative diagram in a triangulated category. By the axioms of a triangulated category, this may be 'completed' to $$ \begin{array}{rccccl} A&\to& B&\to& C&\to\\ \downarrow & &\downarrow&&\downarrow\\ A'&\to& B'&\to& C'&\to \end{array} $$ where both rows are exact triangles. Let $$ \begin{array}{rccccl} A&\to& B&\to& \tilde C&\to\\ \downarrow & &\downarrow&&\downarrow\\ A'&\to& B'&\to& \tilde C'&\to \end{array} $$ be another 'completion' of the initial diagram such that both rows are exact triangles. One gets that $C\cong \tilde C$ and $C'\cong \tilde C'$. My question is: May these two isomorphism be chosen in a way that the diagram $$ \begin{array}{rcl} C&\overset{\cong}{\to}& \tilde C\\ \downarrow & &\downarrow\\ C'&\overset{\cong}{\to}& \tilde C' \end{array} $$ commutes?

$\endgroup$

1 Answer 1

8
$\begingroup$

Not in general.

For example, let the original commutative square be $$\begin{array}{ccc} X[-1]&\to&0\\ \downarrow & &\downarrow\\ 0&\to&X \end{array}$$ for any non-zero object $X$.

This can be completed to a diagram $$\begin{array}{cccccccc} X[-1]&\to& 0&\to& X&\to &X\\ \downarrow & &\downarrow&&\downarrow&&\downarrow\\ 0&\to& X&\to& X&\to& 0 \end{array}$$ with a completely free choice of the vertical map $X\to X$.

Taking two different completions, where this map is zero in one and the identity in the other, gives an example where the answer to your question is "no".

$\endgroup$
3
  • $\begingroup$ Dear Jeremy Rickard, I am sorry but I don't understand your answer anymore: Isn't only the zero map $X\to X$ a correct completion of the diagram? The identity wouldn't make the diagram commute. Did I understand you correctly? $\endgroup$ Commented Feb 4, 2015 at 12:49
  • $\begingroup$ @jeffrey Sorry, you're quite right. I had the diagram the wrong way round. I think it will be correct when I edit. $\endgroup$ Commented Feb 4, 2015 at 13:08
  • $\begingroup$ @jeffrey I've edited, and I think it's correct now. $\endgroup$ Commented Feb 4, 2015 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .