Let $(\mathcal C, \otimes, I)$ be a symmetric monoidal 2-category, and let $X \in \mathcal C$ be a dualizable object, with dual $X^\vee$, unit $coev: I \to X \otimes X^\vee$, and counit $ev : X^\vee \otimes X \to I$ satisfying the triangle identities. Recall that $X$ is [2-dualizable](https://ncatlab.org/nlab/show/fully+dualizable+object) if the unit and counit each fit into doubly-infinite adjoint strings:


$$(1a) \qquad \cdots \dashv ev^{LL} \dashv ev^L \dashv ev \dashv ev^R \dashv ev^{RR} \dashv \cdots$$


$$(1b) \qquad \cdots \dashv coev^{LL} \dashv coev^L \dashv coev \dashv coev^R \dashv coev^{RR} \dashv \cdots$$


Lurie shows [(Prop 4.2.3)](https://people.math.harvard.edu/%7Elurie/papers/cobordism.pdf#page=92) that to check this, it suffices to check that we have the following two adjunctions:


$$(2) \qquad ev^L \dashv ev \dashv ev^R$$


But I don't fully follow the proof. I also seem to recall seeing somewhere that it alternatively suffices to check the following two adjunctions


$$(3a) \qquad ev \dashv ev^R$$


$$(3b) \qquad coev \dashv coev^R$$


**Example:** Criterion (3) implies that any dualizable object in the $(\infty,2)$-category of presentable stable compactly-generated categories is 2-dualizable as an object in the $(\infty,2)$-category of presentable stable categories.


**Question 1:**. To check $(1)$, why does it suffice to check $(2)$?


**Question 2:** Are $(2)$ and $(3)$ equivalent?


**Question 3:** Is the Example discussed anywhere?


Regarding Question 1, Lurie explains in 4.2.3 and the immediately following 4.2.4 that from $ev^R$ we may construct the *Serre twist* $S : X \to X$ and from $ev^L$ we may construct the *inverse Serre twist* $T : X \to X$, but I don't see why these morphisms are inverse to one another. Moreover, I don't follow the argument that this leads to an infinite adjoint string. Finally, I don't see how this yields an infinite adjoint string for coevaluation.


**Question 4:** Are (1a) and (1b) equivalent?



Checking 2-dualizability
I will try to answer questions 1, 2, and 4. Question 3 is somewhat orthogonal to them, and I don't know the answer.


Clearly (1) implies both (2) and (3).


Suppose that (3) holds. Let me write $\sigma$ for the symmetry on $\mathcal{C}$, and consider the map $coev' := \sigma \circ ev^R : I \to X \otimes X^\vee$ and $ev' := coev^R \circ \sigma : X^\vee \otimes X \to I$. The zorro equations for $coev, ev$ imply zorro equations for $coev', ev'$. In other words, $coev'$ and $ev'$ select another 1-duality between $X$ and $X^\vee$.


On the other hand, given $X$, the space of duality data $(X^\vee, coev, ev)$ is contractible [I will only use that it's connected] if it is nonempty. Thus, there is an automorphism of $X^\vee$ which conjugates $coev$ to $coev'$ and $ev$ to $ev'$. Of course, this automorphism is precisely the *Serre*, but I don't really care what its formula is, only that it exists. (Anyway, you can work out a formula: you just use the only formula that identifies different duals with each other.)


Isomorphisms are in particular adjunctible. In particular, since $coev'$ is $coev$ composed with some isomorphisms, they have the same amount of adjunctibility. But $coev$ is by assumption right-adjunctible, whereas $coev' = \sigma \circ ev^R$ is by construction left-adjunctible. So both $coev$ and $coev'$ are both-sides-adjunctible, and ditto for $ev$ and $ev'$.


Now you repeat the game. For example, $coev'$ and $ev'$ admit right adjoints $(coev')^R$ and $(ev')^R$, from which you can build yet another duality $coev'', ev''$ between $X,X^\vee$, and from it conclude that all the players in the game are twice-right-adjunctible and twice-left-adjunctible. Keep going to conclude that (3) implies (1).




---


To finish, I need to show that (2) implies (3). More generally, I want to show the following. Suppose I have a dual pair $(X,X^\vee, coev, ev)$, and I know that $ev : X^\vee \otimes X \to I$ admits a left adjoint $ev^L : I \to X^\vee \otimes X$. Then I claim that $coev : I \to X \otimes X^\vee$ admits a right adjoint $coev^R : X \otimes X^\vee \to I$.


There is only one possible guess for the formula for $coev^R$:
$$ coev^R := (ev \otimes ev) \circ (\sigma \otimes \sigma) \circ (X^\vee \otimes ev^L \otimes X).$$
[Notation: I am using the same name for an object and for its identity 1-morphism.]
The adjunction $ev^L \dashv ev$ consists of 2-morphisms $I \Rightarrow ev \circ ev^L$ and $ev^L \circ ev \Rightarrow X^\vee \otimes X$. I want to supply 2-morphisms $I \Rightarrow coev^R \circ coev$ and $coev \circ coev^R \Rightarrow X \otimes X^\vee$.


But the zorro relation for $ev,coev$ [and some interchanges] provides an isomorphism
$$coev^R \circ coev \cong ev \circ ev^L$$
and I already have a 2-morphism $I \Rightarrow ev \circ ev^L$, so just compose it with my zorro-provided isomorphism to get the desired $I \Rightarrow coev^R \circ coev$.


On the other hand,
$$ coev \cong (X \otimes ev \otimes X^\vee) \circ (coev \otimes coev),$$
and so
$$ coev \circ coev^R \cong [(X \otimes ev \otimes X^\vee) \circ (coev \otimes coev)] \circ [(ev \otimes ev) \circ (\sigma \otimes \sigma) \circ (X^\vee \otimes ev^L \otimes X)].$$
Now apply some interchanges to bring $ev, ev^L$ next to each other:
$$ \cong (X \otimes ev\sigma \otimes ev\sigma \otimes X^\vee) \circ (\sigma \otimes ev^Lev \otimes \sigma) \circ (X \otimes coev \otimes coev \otimes X^\vee)$$
But $ev^Lev$ emits a 2-morphism to $X^\vee \otimes X$, so my complicated composition emits a 2-morphism to
$$ \Rightarrow (X \otimes ev\sigma \otimes ev\sigma \otimes X^\vee) \circ (\sigma \otimes X^\vee \otimes X \otimes \sigma) \circ (X \otimes coev \otimes coev \otimes X^\vee),$$
which, by some zorro and interchange laws, is isomorphic to
$$ \cong X \otimes X^\vee.$$


The last step, which I'll leave to you to enjoy, is to check that indeed the 2-morphisms I just constructed do supply an adjunction $coev \dashv coev^R$.


