Tensor product of mapping cones

Question

Fix a ring $R$. If $A^*_i \to B^*_i \to C^*_i \to A^*_i[1]$ is a distinguished triangle of complexes of $R$-modules, for $i=1$ and $2$ (so $C_i^* = cone(f_i^*)$ where $f_i^*: A_i^*\to B_i^*$), is there a nice way to express the derived tensor product $C^*_1 \otimes^L C^*_2$ in terms of the above morphisms $A^*_i \to B^*_i$?

Answer 1

Th answer by White is not quite right, even for short exact sequences of vector spaces (for dimension reasons).

One way to describe $C_1 \otimes^L C_2$ is via filtrations. Regard each of the given exact triangles as defining an increasing filtration of length $1$ on $B_i$ with $\mathrm{gr}_0 = A_i$ and $\mathrm{gr}_1 = C_i$. Then $B_1 \otimes^L B_2$ inherits an increasing filtration of length $2$ with $\mathrm{gr}_0 = A_1 \otimes^L A_2$, $\mathrm{gr}_1 = \big(A_1 \otimes^L C_2\big) \oplus \big(C_1 \otimes^L A_2\big)$, and $\mathrm{gr}_2 = C_1 \otimes^L C_2$.

Answer 2

Yes, there is a nice way. The mapping cone $C_i^*$ is the homotopy pushout of $\ast \gets A_i^* \to B_i^*$, and $-\otimes X$ is a monoidal left Quillen functor (symmetric if the underlying ring is commutative), so $-\otimes^L -$ commutes with homotopy colimits. Hence, $C_1^* \otimes^L C_2^*$ is the mapping cone of $A_1^* \otimes^L A_2^* \to B_1^* \otimes^L B_2^*$ . Derived functors and homotopy colimits can be computed via projective resolutions, so you have very classical inductive formulas to compute these quantities if you like.

EDIT: The original answer is too simplistic, so this edit fixes it (thanks Thomas for pointing this out). Instead of identifying $C_1^* \otimes^L C_2^*$ as "the mapping cone", it should say "$C_1^* \otimes^L C_2^*$ is the homotopy colimit of a cube built from the maps $A_i^* \to B_i^*$". My original answer was ignoring the grading, because I had just posted an answer, with the same "homotopy colimit" idea, to a simpler question.