Let us call a commutative square $$ \require{AMScd} \begin{CD} A @>{g'}>> B \\ @V{f'}VV @VV{f}V \\ C @>>{g}> D \end{CD} $$ in a triangulated category split homotopy cartesian if the ("split") triangle $A\stackrel{f'\bigoplus g'}{\to}C\bigoplus B\stackrel{g\bigoplus -f}{\to} D \stackrel{0}{\to} A[1]$ is distinguished. Certainly, this is equivalent to the existence of an isomorphism $C\bigoplus B\cong A\bigoplus D$ that is "coherent" with these arrows. Yet do you know any criteria that would be "closer to the original square"?
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asked Nov 11, 2022 at 11:25
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2$\begingroup$ This sounds like what is called a (homotopy) pushout square in stable infinity-category (or model category). The first chapter of Higher Algebra by Lurie should be relevant. (Topologist like e.g. the example of the singular chains on $U\cap V$, $U$, $V$ and $U\cup V$ for an open cover.) $\endgroup$– Lennart MeierCommented Nov 11, 2022 at 11:40
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3$\begingroup$ If you insist on triangulated categories (as opposed to stable $\infty$-categories, as Lennart rightfully suggests), this kind of things (at least the possibly non-splitted version) is discussed in Neeman's book Triangulated Categories (Annals of Math. Studies 148, Princeton Univ. Press) in Chapter 1, §1.4, pp. 52-60. The fact that these define "weak pullbacks" is discussed in Chapter 6, Example 1.6.2, page 184. $\endgroup$– D.-C. CisinskiCommented Nov 11, 2022 at 12:12
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3$\begingroup$ The splitting condition seems to indicate that these actually are pullbacks (and pushouts)... $\endgroup$– D.-C. CisinskiCommented Nov 11, 2022 at 12:17
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2$\begingroup$ Thank you! Yes, they are pullbacks and pushouts. Possibly, this will help me to find something useful. $\endgroup$– Mikhail BondarkoCommented Nov 11, 2022 at 12:41
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