# How can I functorially dualise in a symmetric monoidal $(\infty,1)$-category with duals?

If $\mathcal{C}$ is a symmetric monoidal $(\infty,1)$-category with duals, then there should be a functor $$d: \mathcal{C} \longrightarrow \mathcal{C}^{op}$$ such that $d(x)$ is dual to $x$ for all objects $x \in \mathcal{C}$. How can one construct such a functor?

Of course, the above is only a very weak formulation of what one would actually expect. In fact, it should be possible to choose $d$ such that there is a coherent homotopy $d^{op} \circ d \simeq \operatorname{Id}_{\mathcal{C}}$. By 'coherent' I mean that $\mathcal{C}$ should have homotopy fixed-point data for the involution $$(\ )^{op}: \mathit{Cat}_\infty^{\otimes} \longrightarrow \mathit{Cat}_\infty^{\otimes}.$$ And while I'm on it, let me give the most general version of the quesion:

Is there a canonical (or maybe even essentially unique) trivialisation of the $(\ )^{op}$ involution when restricted to the full subcategory $\mathit{Cat}_\infty^{\otimes,d}\subset \mathit{Cat}_\infty^{\otimes}$ of those symmetric monoidal $(\infty,1)$-categories in which every object is dualisable?

I am aware of a solution to this question in the $1$-categorical context; it goes as follows:

For a symmetric monoidal $1$-category $\mathcal{C}$ one constructs a category $D(\mathcal{C})$ where objects are tuples $(x,y,e,c)$ with $x,y$ two objects of $\mathcal{C}$ and $(e,c)$ is an evaluation-coevaluation pair that exhibits them as dual. The morphisms $\varphi:(x,y,e,c) \to (x',y',e',c')$ in this category are pairs $(f:x \to x',g:y' \to y)$ such that $f$ is dual to $g$ with respect to the duality data specified. This category admits canonical projections $$\mathcal{C} \longleftarrow D(\mathcal{C}) \longrightarrow \mathcal{C}^{op}.$$ If $\mathcal{C}$ has duals, one can use essential uniqueness of duals to show that both projections are categorical equivalences. Using this construction it should be rather straight-forward to show all claims made above for the special case of $1$-categories. (Except for uniqueness of the trivialisation.)

Now the problem is that I have no idea how to generalise this proof to the $\infty$-categorical situation. For instance, for the two morphisms $f$ and $g$ to be dual would then be additional structure instead of a property.

I would be happy to use the cobordism hypothesis in dimension $1$, if that is of any help. (It can be used to construct $d$ on the maximal subgroupid $\mathcal{C}^\sim \subset \mathcal{C}$, but it seems hard to use it to say anything about duals of non-invertible morphisms. As a side-note: Where can I find a proof or even a proof sketch of the cobordism hypothesis in dimension $1$?)

I'm interested in answers at any stage of generality.

• Side note: Cobordism hypothesis in dimension 1 – Noah Snyder Jul 18 '18 at 3:10
• For the question with only invertible morphisms, see Higher Algebra, Lemma 4.6.1.10 (there $D(C)$ has morphisms $g\colon y\rightarrow y'$). – Pavel Safronov Jul 18 '18 at 6:56
• Roughly one wants this to follow from some facts about adjoint functors. That is, $X^*$ is characterized by $X^* \otimes$ being adjoint to $X \otimes$ (note: by unitality the functor determines the object). So you just want to know that taking adjoints is functorial. Roughly this comes from adjoints being "unique", i.e. that the space of them is contractible. This is closely related to the fact that adjoints can always be made "coherent adjoints," but I'm not comfortable enough with the details to try to turn this sketch into an answer. – Noah Snyder Jul 20 '18 at 17:45

One way to construct the duality functor ${\cal C} \to {\cal C^{\rm op}}$ is through the notion of a pairing of $\infty$-categories (see HA, Definition 5.2.1.5). In particular, in this case we're talking about a self-pairing on ${\cal C}$, which by definition is a right fibration $\mu:{\cal M} \to {\cal C} \times {\cal C}$, classified by a functor $b: {\cal C}^{\rm op} \times {\cal C}^{\rm op} \to {\cal S}$ to spaces. We say that $\mu$ is left representable if for every $x \in {\cal C}$ the functor $y \mapsto b(x,y)$ is representable in ${\cal C}$, and right representable if for every $y \in {\cal C}$ the functor $x \mapsto b(x,y)$ is represetable in ${\cal C}$. If $\mu$ is a pairing which is both left and right representable then it determines an adjunction between ${\cal C}$ and ${\cal C}^{\rm op}$. We say that $\mu$ is a perfect pairing if it is both left and right representable and the associated adjunction is an equivalence. One can then show that the data of a perfect pairing $\mu: {\cal M} \to {\cal C} \times {\cal C}$ is in fact equivalent to the data of an equivalence ${\cal C} \to {\cal C}^{\rm op}$. The advantage of this point of view is that the $\mathbb{Z}/2$-action on the space of equivalences ${\cal C} \to {\cal C}^{\rm op}$ becomes explicit: it is given by sending $\mu: {\cal M} \to {\cal C} \times {\cal C}$ to ${\rm swap} \circ \mu: {\cal M} \to {\cal C} \times {\cal C}$, where ${\rm swap}:{\cal C} \times{\cal C} \to {\cal C} \times {\cal C}$ is the equivalence that swaps the two components. In particular, to construct an equivalence ${\cal C} \stackrel{\simeq}{\to} {\cal C}^{\rm op}$ which is self-dual in a homotopy coherent manner is equivalent to constructing a perfect pairing ${\cal M} \to {\cal C} \times {\cal C}$ together with a $\mathbb{Z}/2$-action on ${\cal M}$ which lifts the swap action on ${\cal C} \times {\cal C}$. Note that this swap action can itself be encoded as a right fibration $p:{\rm Sym}({\cal C}) \to {\rm B}(\mathbb{Z}/2)$, where ${\rm B}(\mathbb{Z}/2)$ is the groupoid with one object whose endomorphism group is $\mathbb{Z}/2$, and the fiber of $p$ over this one object is ${\cal C} \times {\cal C}$. One can then encode a lift of the swap action to ${\cal M}$ by a right fibration $\widetilde{\cal M} \to {\rm Sym}({\cal C})$ whose restriction to ${\cal C} \times {\cal C}$ is ${\cal M}$. In general, we may refer to right fibrations $\widetilde{\cal M} \to {\rm Sym}({\cal C})$ as symmetric pairings, and say that a symmetric pairing is perfect when its base change to ${\cal C} \times {\cal C} \subseteq {\rm Sym}({\cal C})$ is perfect. The notion of a perfect symmetric pairing then encodes a duality on ${\cal C}$, i.e., a self-dual equivalence $d:{\cal C} \to {\cal C}^{\rm op}$ (with all the higher coherences taken into account).

Now in the case of a symmetric monoidal $\infty$-category with duals, the pairing encoding the duality is the right fibration ${\cal M} \to {\cal C} \times {\cal C}$ classified by the functor $b_{\cal C}: {\cal C} \times {\cal C} \to {\cal S}$ sending $(x,y)$ to ${\rm Map}_{\cal C}(x \otimes y,1_{\cal C})$. There is a direct way to construct this as a symmetric perfect pairing. Indeed, suppose that $\pi:{\cal C}^{\otimes} \to {\rm Fin}_*$ is the coCartesian fibration encoding the symmetric monoidal structure on ${\cal C}$, and let ${\cal C}^{\otimes}_{\rm act} \to {\rm Fin}$ be its active part, i.e., the base change to the category ${\rm Fin}$ of finite sets via the functor $I \mapsto I_+$. Then we can identify ${\rm B}(\mathbb{Z}/2)$ as the subcategory of ${\rm Fin}$ consisting of the object $\left<2\right>^{\circ} = \{1,2\} \in {\rm Fin}$ and all its automorphisms, and the base change of ${\cal C}^{\otimes}_{\rm act}$ to ${\rm B}(\mathbb{Z}/2) \subseteq {\rm Fin}$ is naturally equivalent to ${\rm Sym}({\cal C})$. If $1_{\cal C} \in ({\cal C}^{\otimes}_{\rm act})_{\left<1\right>^{\circ}}$ now denotes the unit then the right fibration $$({\cal C}^{\otimes}_{\rm act})_{/1_{\cal C}}\times_{{\rm Fin}} {\rm B}(\mathbb{Z}/2) \to {\cal C}^{\otimes}_{\rm act} \times_{{\rm Fin}}{\rm B}(\mathbb{Z}/2) \simeq {\rm Sym}({\cal C})$$ is a symmetric pairing whose underlying pairing is $b_{\cal C}$ above. When ${\cal C}$ has duals this symmetric pairing is perfect, yielding the associated self-dual equivalence $d:{\cal C} \to {\cal C}^{\rm op}$.

I don't consider this an actual answer to the question, as it uses a hypothesis that is, to my knowledge, not yet proven. I hope that there is a simpler answer to this question that is more elementary and does not rely on such an assumption.

I've come up with the following argument and, even though it does not answer the question, I feel like it would be helpful to share it. The argument has the following downsides:

• It is rather lengthy when spelled out in detail. For now I will only provide a sketch, but I'm happy to fill in details if requested.

• It constructs $d$ only after forgetting the symmetric monoidal structure. (This part is not actually hard to fix.)

• It relies on the following hypothesis:

Hypothesis: We have a model $B_1([n])$ of the symmetric monoidal $\infty$-category with duals freely generated by the poset $[n]$ that we understand well enough to give an isomorphism $$\rho_n: B_1([n]) \cong B_1([n]^{op}).$$ (There also is a canonical isomorphism $B_1([n]^{op}) \cong (B_1([n]))^{op}$. After composing with this $\rho_n$ should send objects to their duals.)

Assuming the $1$d-cobodism hypothesis with sigularities such a model can be given in terms of a category of framed one-dimensional bordisms with certain labels. (See this question.) In the specific model I have in mind $\rho_n$ can be defined by simply reversing the orientation.

However, I'm not aware of a complete proof of the cobordism hypothesis with singularities, even in dimension $1$. If someone is, I'd be very glad to here about it.

Sketch of the argument:

Let's fix some symmetric monoidal $\infty$-category with duals $\mathcal{C}^\otimes$ and write $\mathcal{C}$ for its underlying $\infty$-category. Now let $X: \Delta^{op} \to \mathit{Spaces}$ denote the simplicial space that represents $\mathcal{C}$ as a complete Segal space. By definition this is given by $$X_{[n]} := (\mathit{Fun}_\infty( N[n], \mathcal{C}) )^\sim.$$ (Here the $(\ )^\sim$ takes the underlying $\infty$-groupoid of the functor category and interprets it as a space. The functor $N$ is the inclusion $\Delta \to \mathit{Cat} \to \mathit{Cat}_\infty$. )

By our hypothesis the symmetric monoidal $\infty$-category $B_1([n])$ is free with duals on $[n]$, so there is an equivalence of spaces $$Y_{[n]}:= (\mathit{Fun}_\infty^\otimes( B_1([n]), \mathcal{C}^\otimes) )^\sim \ \simeq \ (\mathit{Fun}_\infty( N[n], \mathcal{C}) )^\sim =X_{[n]}$$

Hence $Y_{\bullet}: \Delta^{op} \to \mathit{Spaces}$ defines a Segal space equivalent to $X$ and therefore also models the $\infty$-category $\mathcal{C}$. By our hypothesis we have $\rho_n:B_1([n])\cong B_1([n]^{op})$. Assuming that these isomorphisms play well with the fuctoriality in $[n]$ this induces an equivalence $$d: \mathcal{C} \sim Y_\bullet \cong (Y_\bullet)^{op} \sim \mathcal{C}^{op}.$$

One can now check that the assumptions of $\rho_n$ imply that $d$ sends objects and morphisms to their duals.

With a bit more care $d$ can be made symmetric monoidal and functorial in $\mathcal{C}$ such that it indeed gives a canonical trivialisation of $\ ^{op}$ on $\mathit{Cat}_\infty^{\otimes,d}$. I don't see how one could use this approach to show that the trivialisation is essentially unique.