By the work of Verdier, we know that cones in a triangulated category $\mathcal{T}$ are functorial if and only if $\mathcal{T}$ is semisimple abelian. However, in these notes, it is said that

In the context of triangulated categories, it is well known that cones are not functorial. However, we have just proven that if a triangulated category arises as the homotopy category of a stable $\infty$-category, cones in it are indeed functorial at the $\infty$-level.

I must admit that I don't understand what it means that cones are functorial at the $\infty$-level. Since the derived category of an abelian category (with enough injectives) is the homotopy category of the derived infity category, which is a stable $\infty$-category, that would mean that we have functorial cones in derived abelian categories. What does that mean? Can I "take kernels"?


1 Answer 1


Let $\mathcal{C}$ be a stable $\infty$-category. Then $\mathcal{C}$ has a homotopy category $h \mathcal{C}$, which is triangulated. The collection of morphisms $f: X \rightarrow Y$ of $\mathcal{C}$ can be organized into an $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$. The operation $f \mapsto \mathrm{Cone}(f)$ can be obtained from a functor of $\infty$-categories $\mathrm{Fun}( \Delta^1,\mathcal{C} ) \rightarrow \mathcal{C}$. You can pass to homotopy to get a functor of ordinary categories $$\mathrm{Cone}: h \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow h \mathcal{C}.$$

The functor $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$ is equipped with an evaluation functor $e: \Delta^1 \times \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow \mathcal{C}$. You can take homotopy categories here, to get a functor of ordinary categories $[1] \times h\mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow h\mathcal{C}$, which can be identified with a functor (again of ordinary categories) $$ U: h \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow \mathrm{Fun}( [1], h\mathcal{C} ).$$ Here $[1]$ denotes the category $\{ 0 < 1 \}$ consisting of two objects and one morphism between them.

The phenomenon you're asking about is due to the fact that $U$ is not an equivalence of categories. Moreover, the functor $\mathrm{Cone}$ does not factor through $U$. If $f: X \rightarrow Y$ and $f': X' \rightarrow Y'$ are morphisms of $\mathcal{C}$, then morphisms from $f$ to $f'$ on in the $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$ can be thought of triples $(u,v,h)$, where $u: X \rightarrow X'$ and $v: Y \rightarrow Y'$ are morphism of $\mathcal{C}$ and $h$ is a homotopy from $f' \circ u$ to $v \circ f$. In these terms, the functor $U$ is given on morphisms by the construction $[(u,v,h)] \mapsto ( [u], [v] )$ where $[s]$ denotes the homotopy class of a morphism $s$. In particular $U$ "forgets" the data of the homotopy $h$, and fails to be a faithful functor.


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