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It is a standard fact that smashing with a fixed spectrum $Z$ preserves cofiber sequences. So if I have a cofiber sequence $$X \xrightarrow{f} Y \rightarrow C_f$$ then there is also a cofiber sequence $$Z \wedge X \rightarrow Z \wedge Y \rightarrow Z \wedge C_f$$

If more generally I have a map $Z \xrightarrow{g} W$, is there any formula for the cofiber of the map $$Z \wedge X \xrightarrow{g \wedge f} W \wedge Y$$ in terms of $C_f$ and $C_g$? (The above discussion corresponding to $g = \mathrm{id}_Z$).

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up vote 7 down vote accepted

By factoring $g\wedge f$ as $X\wedge f$ followed by $g\wedge Y$ you see that it's in the middle of a cofiber sequence $Z\wedge C_f\to \ ?\to C_g\wedge Y$. Similarly it's in the middle of a cofiber sequence $C_g\wedge X\to\ ?\to W\wedge C_f$. It's also in the middle of a cofiber sequence $(Z\wedge C_f)\vee (C_g\wedge X)\to \ ?\to C_g\wedge C_f$. We could probably come up with more. But no formula in the sense that I think you mean. It may be instructive to think of the special case where $X$ and $W$ are both contractible.

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Yes, I somehow figured that a set formula was too much to ask for. But thanks for the tips. I will try these in my situation and see if I can get anything to fall out. – Eric Finster Sep 7 '10 at 19:39

An axiomatization of exactly how smash products of cofiber sequences should behave is given in the context of triangulated categories in my paper

The additivity of traces in triangulated categories. Advances in Mathematics 163(2001), 34--73.

Of course, the axioms hold in the motivating example of the homotopy category of spectra.

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