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Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural map $$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ My question is:

Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?

i.e. is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$

If this isn't a "universal triangulated category", does there exist such a construction?

Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural functor $$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ My question is:

Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?

That is, is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$

If this isn't a "universal triangulated category," does there exist such a construction?

Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural map $$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ My question is:

Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?

i.e. is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$

Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural map $$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ My question is:

Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?

i.e. is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$

If this isn't a "universal triangulated category", does there exist such a construction?

Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the associated simplicial category $s\mathcal{C}$ (which is an $\infty$-category) and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of an $\infty$-category). Then we get a natural map $$\mathcal{C}\rightarrow \text{Ho}(s\mathcal{C}).$$ An example of this construction would be to set $\mathcal{C}=A$ for some abelian category, in which case by Dolb-Kan, we have $\text{Ho}(s\mathcal{C})\cong D(A).$ My question is:

Does $\text{Ho}(s\mathcal{C})$ satisfy some universal property?

i.e. is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(s\mathcal{C})?$

Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural map $$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ My question is:

Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?

i.e. is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$