For a diagram category $\Gamma$ and and a cocomplete category $\mathcal{C}$, we have an equivalence $$\mathrm{Fun}(\Gamma,\mathcal{C}) \simeq \mathrm{Adj}(Set^{\Gamma^{Op}},C)$$ where for $F: \Gamma \to C$ we have a pair of adjoint functors $\mathcal{L}_F: Set^{\Gamma^{Op}} \to \mathcal{C}: S \mapsto \mathrm{colim}_{h_\gamma \to S}F(\gamma)$ (left adjoint) and $\mathcal{N}_F: \mathcal{C} \to Set^{\Gamma^{Op}}: c \mapsto \mathcal{C}(F(-),c)$ (right adjoint).

For $\Gamma = \Delta$, this states that a cosimplicial object in $\mathcal{C}$ defines an adjunction between simplicial sets and $\mathcal{C}$. For example, when $\mathcal{C}=Top$ and the cosimpicial object consists of standard toplogical simplices, $\mathcal{L}_F$ is geometric realization and $\mathcal{N}_F$ is singular set.

Generally for $\Gamma=\Delta$, the explicit formula for $\mathcal{L}_F$ is $$\mathcal{L}_F(S_\bullet) = Coeq(\amalg_{\phi:[m]\to[n]}F([m])\times S_n \rightrightarrows \amalg_{[n]}F([n])\times S_n).$$

When instead of a simplicial set $S_\bullet$ we have a cosimplicial category $A^\bullet$, there exists a (somewhat dual) notion of totalization: $$\mathrm{Tot}(A^\bullet):=Eq(\Pi_{[n]}\mathrm{Fun}(Iso(n),A^n) \rightrightarrows \Pi_{[n] \to [m]}\mathrm{Fun}(Iso(n),A^m))$$

where $Iso(n)$ stands for the category "string of $n$ isomorphisms". This totalization is a correct notion (e.g. it works well for $A^\bullet$ being sheaves on a Čech nerve of a cover).

I would expect it to be a part of some general picture similar to $\mathrm{Fun}(\Gamma,\mathcal{C}) \simeq \mathrm{Adj}(Set^{\Gamma^{Op}},C)$, only in enriched setting (sets are replaced by categories) and for $\Gamma=\Delta^{Op}$. However, it is not exactly the picture above, as $Iso(n)$ is a functor $\Delta \to Cat$ not $\Delta^{Op} \to Cat$.

What is this general picture for totalizations? Is there an adjoint functor?


There are a number of ways of expressing this duality.

  • The colimit in the question may be written as the coend $\mathcal L_F(S_\bullet) = \int^{[n] \in \Delta} F([n]) \times S_n$, while the limit may be written as $\mathrm{Tot}(A^\bullet) = \int_{[n] \in \Delta} \mathrm{Hom}(Iso([n]), A^n))$.

  • The colimit in the question is the weighted colimit $\mathcal L_F(S_\bullet) = F \otimes S$ (the colimit "of" the functor $S$, "weighted" by the presheaf $F$). The limit in the question is the weighted limit $\mathrm{Tot}(A^\bullet) = \{Iso,A\}$ (the limit "of" the functor $A$, "weighted" by the copresheaf $A$).

  • The colimit in the question computes the the left Kan extension $\mathcal L_{(-)}(S_\bullet)$ of the functor $S_\bullet$ along the Yoneda embedding $\Delta \to Fun(\Delta^{op},\mathsf{Set})$. The limit in the question (generalized from the case of the functor $Iso$ in the obvious way) computes the right Kan extension $\mathrm{Tot}^{(-)}(A^\bullet)$ of the functor $A^\bullet$ along the coYoneda embedding $\Delta \to Fun(\Delta,\mathsf{Set})^{op}$.

  • The category of presheaves $Fun(\Delta^{op},\mathsf{Set})$ is the free cocompletion of $\Delta$, so that $Fun(\Delta \simeq Fun^{cocts}(Set^{\Delta^{op}},\mathcal C)$ for cocomplete $\mathcal C$ (where $Fun^{cocts}$ means cocontinuous functors). Dually, the opposite of the category copresheaves $Fun(\Delta,\mathsf{Set})^{op}$ is the free completion of $\Delta$, i.e. there is an equivalence $Fun(\Delta,\mathcal{C}) \simeq Fun^{cts}((Set^{\Delta})^{op},\mathcal{C})$ (where $Fun^{cts}$ means continuous functors).

  • Applying the adjoint functor theorem to the previous point, we obtain the equivalence mentioned in the question $Fun(\Delta,\mathcal C) \simeq LAdj(Set^{\Delta^{op}},\mathcal C)$ (the cateogry of left adjoint functors). Dually, we have $Fun(\Delta,\mathcal{C}) \simeq RAdj((Set^{\Delta})^{op},\mathcal{C})$ (the category of right adjoint functors.

And all of this works for any small $\Gamma$ in place of $\Delta$, and any complete and cocomplete $\mathcal{C}$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.