# Grothendieck Construction, Categories of Operators and Opposites

Given a symmetric monoidal category $C$, we can construct its endomorphism operad (or multicategory) $End(C)$ whose objects are the objects of $C$, and for which the multimorphisms from $\{c_1,\ldots,c_n\}$ to $c$ are given by the hom set $Hom_C(c_1\otimes\cdots c_n,c)$. From this symmetric operad we can then produce a category of operators whose objects are finite lists of objects of $C$ and whose morphisms are described (roughly) in the following way: given two lists $\{c_1,\ldots,c_n\}$ and $\{d_1,\ldots,d_m\}$, a morphism between them is a morphism $\phi:\langle n\rangle\to\langle m\rangle$ in finite pointed sets (the category often denoted $\Gamma$ by Segal and others, and $\mathcal{F}in_\ast$ by Lurie) and for each element $i\in\langle m\rangle$ and set $\{j\in\phi^{-1}(i)\}$ a multimorphism of $End(C)$ going from $\{c_j\}$ to $d_i$. Let's denote the application of these procedures to $C$ by $End(C)^\otimes$. Essentially by construction, $End(C)^\otimes$ admits a Grothendieck opfibration to $\Gamma$, and as such, corresponds to a pseudofunctor $F_C:\Gamma\to Cat$. One might even say that this pseudofunctor defines the symmetric monoidal structure on $C$.

Given a symmetric monoidal category there is a symmetric monoidal structure on $C^{op}$, so we may also consider $End(C^{op})^\otimes$. Similarly, we obtain a pseudofunctor $F_{C^{op}}:\Gamma\to Cat$.

The sort of theorem I'm after is the following:

The composition of pseudofunctors $\Gamma\overset{F_{C}}\to Cat\overset{op}\to Cat$ is equivalent to the pseudofunctor $\Gamma\overset{F_{C^{op}}}\to Cat$.

Or at the very least, I just want to say that if I apply the Grothendieck construction $\Gamma\overset{F_{C}}\to Cat\overset{op}\to Cat$ I get a symmetric monoidal category which is equivalent, as a symmetric monoidal category, to $C^{op}$ with the opposite symmetric monoidal structure.

• Hi Jon, I'd say that both pseudo-functors are actually equal. The opposite of a product of categories is the product of opposite categories, and $F_C$ is given by $F_C(\langle n\rangle)=C^n$. – Fernando Muro Jun 14 '16 at 18:36

• So what was wrong with Fernando's argument (simply unpacking the definitions of $\mathrm{op} \circ F_C$ and $F_{C^{\mathrm{op}}}$ and observing that they coincide)? – Najib Idrissi May 23 '19 at 9:39