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Assume that C is a stable infinity category; $SH_{fin}$ is the homotopy category of finite spectra. Is there a canonical bi-functor (action? module structure?) $SH_{fin}\times hC \to hC$?

Is there any "canonical" formulation of this statement; what about the properties of this bi-functor? It possibly follows from Proposition 4.8.12.8 of Lurie's Higher Algebra, but I don't really understand what the latter says and do not want to read the whole book.

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2 Answers 2

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Yes: Since $\mathcal{C}$ is stable, $\operatorname{Fun}(\mathcal{C},\mathcal{C})$ is stable, too. In particular, it has finite colimits, so $\operatorname{Ind}\operatorname{Fun}(\mathcal{C},\mathcal{C})$ has all colimits. So we get a unique colimit-preserving functor $\mathcal{S} \to \operatorname{Ind}\operatorname{Fun}(\mathcal{C},\mathcal{C})$ taking the one-point space to the identity functor, and this extends further to a unique exact functor $\operatorname{Sp}\to \operatorname{Ind}\operatorname{Fun}(\mathcal{C},\mathcal{C})$. Passage to compact objects gives a functor $\operatorname{Sp}^\omega \to \operatorname{Fun}(\mathcal{C},\mathcal{C})$, or equivalently $\operatorname{Sp}^\omega \times \mathcal{C}\to \mathcal{C}$.

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The point is simply that every finite spectrum $X$ is a finite colimit of copies of the sphere spectrum and $0$s, and for $c$ in $C$ you just take the same colimit of copies of $c$ and $0$s to define $X\otimes c.$ For instance, $S^1\otimes c$ is the suspension of $c,$ ie the pushout of $0\leftarrow c\to 0$, and so on.

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  • $\begingroup$ Of course, this isn’t anything like a full proof, just the core idea as a complement to the other answer. $\endgroup$ Commented Jul 7, 2023 at 15:45

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