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?

$\endgroup$
5
$\begingroup$

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".

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user8463524 Feb 4 '15 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. $\endgroup$ – Jeremy Rickard Feb 4 '15 at 13:08
  • $\begingroup$ @jeffrey I've edited, and I think it's correct now. $\endgroup$ – Jeremy Rickard Feb 4 '15 at 13:15

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.