# Relationship of additive and triangulated structures in the triangulated cateogry

In the triangulated category $T$, say we have two distinguished triangles: $$X \xrightarrow{f} Y \rightarrow C_f \rightarrow X[1],$$ $$X \xrightarrow{g} Y \rightarrow C_g \rightarrow X[1]$$ Of course these data uniquely determines the triangele $$X \xrightarrow{f+g} Y \rightarrow C_{f+g} \rightarrow X[1]$$ The natural question is if one can write down some kind of "reasonable" relationship between $C_{f+g}$ and $C_{f}$ and $C_{g}$? (seems like natural approach $f+g=\triangledown \circ (f\oplus g) \circ \triangle$ does not lead to anything reasonable. Maybe the fact that cone construction is not functorial, does not allow the nice interplay between additive and triangulated structures)

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Probably the answer is 'no', but how to prove it? – Fernando Muro Apr 18 '13 at 16:21
The prototype for the notion of "mapping cone" is the notion of cokernel, so maybe it would be helpful to ask yourself: given two maps between, say, abelian groups, is their a relationship between the cokernel of the sum and the cokernel of the indiviudal maps? – Dylan Wilson Apr 18 '13 at 17:21
**is there. (whoops) – Dylan Wilson Apr 18 '13 at 17:22
@Dylan, in my experience, the analogy of mapping cones with cokernels is usually misleading. – Fernando Muro Apr 19 '13 at 2:23
Assuming the category is $k$-linear you can do a more general thing by taking the cone of $\lambda f + \mu g$ for $\lambda,\mu \in k$. It gives a family of objects specializing to $C_f$, $C_g$, and $C_{f+g}$. So, one can say that all these objects are in the same deformation family. – Sasha Apr 19 '13 at 17:59