I have a $\otimes$-triangulated category $\mathcal T$ and two triangles in $\mathcal T$: $$ x_0\to x_1\to c_x\to \Sigma x_0\ \ \ \text{and}\ \ \ y_0\to y_1\to c_y\to \Sigma y_0. $$ Consider the following diagram, where the object $P$ is constructed as a homotopy push-out of the obvious maps $\Sigma^{-1}c_x\otimes y_0 \leftarrow\Sigma^{-1}c_x\otimes\Sigma^{-1}c_y\to x_0\otimes \Sigma^{-1} c_y$ (one can do this either as in Neeman's book, or assuming that there is some enhancement for $\mathcal T$): enter image description here Is it true that the cone of the map $\varphi$ obtained this way is isomorphic to $x_1\otimes y_1$?

The question is relatively natural, so I hope that somebody with a little more working experience with $\otimes$-triangulated categories will be able to give a simple proof or a known counterexample in some concrete situation. If this is not known in general, I am happy to assume that $\mathcal T$ has some kind of enhancement (e.g., $\mathcal T$ is the homotopy category of a stable $(\infty,1)$-category, or $\mathcal T$ is the base of a strong and stable derivator), or even, to start with, to assume that $\mathcal T$ is the derived category of a commutative ring.

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    $\begingroup$ When you are working in a stable ∞-category, you are looking for the total cofiber that can be computed as in the bonus part of my answer here (technically, that's the total fiber instead, but you just need to dualize the arguments). I suspect that the statement is also true in a generic ⊗-triangulated category, using the octahedral axiom in creative ways. $\endgroup$ – Denis Nardin Aug 2 '19 at 20:50
  • $\begingroup$ @DenisNardin That is very nice, I like this way to see the problem! I also believe that this should be true in general $\otimes$-triangulated categories, but I just need the proof when $\mathcal T$ is the base of a strong and stable, monoidal derivator. I believe that your argument will be not too complicated to translate in that context. If you want to make your comment into an answer I'll be happy to accept it. $\endgroup$ – Simone Virili Aug 2 '19 at 21:03

In what follows I am going to assume $\mathcal{T}$ is the homotopy category of a stably symmetric monoidal ∞-category. The argument I am going to give can probably be generalized to any case of interest, but I'll stay in familiar territory (for me).

What you are trying to compute is usually called the total cofiber of the square

$$\require{AMScd} \begin{CD} \Sigma^{-1}c_x \otimes \Sigma^{-1}c_y @>>> x_0 \otimes \Sigma^{-1}c_y\\ @VVV @VVV\\ \Sigma^{-1}c_x\otimes y_0 @>>> x_0\otimes y_0 \end{CD}$$

Now the total cofiber of a square can be computed by taking the cofiber of the induced map on the cofiber of any pair of parallel arrows (this is a well known result, at the end of this answer of mine you can find a short proof in the case of the total fiber), so we can compute the required object by first taking the cofibers of parallel arrows. Since the tensor product preserves fiber sequences, and the cofiber of $\Sigma^{-1}c_x\to x_0$ is $x_1$, the required object is the cofiber of $$x_1\otimes \Sigma^{-1}c_y\to x_1\otimes y_0$$ Hence it coincides with $x_1\otimes y_1$, as required.

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    $\begingroup$ Total cofibers also behave nicely in any (stable) derivator, since you're just talking about breaking down homotopy Kan extensions in pieces. I'm skeptical whether you could prove this using just the octahedral axiom...almost certainly you'd at least get stuck trying to generalize to more objects. $\endgroup$ – Kevin Arlin Aug 3 '19 at 0:04
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    $\begingroup$ Yes, actually the proof in the derivator setting is nicely written in Moritz Gorth's book project on derivators (see Thm.12.5.8 in math.uni-bonn.de/~mgroth/monos/intro-to-der-1.pdf) $\endgroup$ – Simone Virili Aug 3 '19 at 11:03

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