If a $\otimes$-idempotent object has a dual, must it be self-dual? Let $C$ be a symmetric monoidal category.


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*Recall that a dual for $X \in C$ is an object $X^\vee$ and maps $\eta: I \to X \otimes X^\vee$ and $\varepsilon: X^\vee \otimes X \to I$ (where $I$ is the monoidal unit) satisfying the triangle identities.

*Let's say that $X$ is self-dual if there is a dual $X^\vee$ for $X$ with $X \cong X^\vee$.

*Let's say that $X$ is idempotent if $X \cong X \otimes X$.

*Let's say that $X$ is well-idempotent if there is a map $i: I \to X$ such that $X \otimes i: X \to X \otimes X$ is an isomorphism.

*Dually, $X$ is co-well-idempotent if there is a map $p: X \to I$ such that $X \otimes p: X \otimes X \to X$ is an isomorphism.
 Clearly, if $X$ is self-dual and idempotent, then $X$ is both well-idempotent and co-well-idempotent. Conversely, I ask
Questions:


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*If $X$ is idempotent and dualizable, then is $X$ self-dual?

*If $X$ is well-idempotent and dualizable, then is $X$ self-dual?

*If $X$ is well-idempotent and co-well-idempotent and dualizable, then is $X$ self-dual?
I suspect the answer to all these questions is "no", but I don't know any examples.
 A: EDIT: It looks like I was too hasty to delete this answer. I think I have fixed the gap that Anton noted in the comments and in his answer. As Anton pointed out in his answer, the trace loop that I dropped out at the end is nontrivial, but it can be removed from any diagram where an "X" appears in another connected component. Intuitively, an idempotent should have dimension 1 wherever it is nonzero. I've oriented the string diagrams vertically so that any horizontal cross-section can be read left-to-right in symbolic order.
The answer to question 2 (hence 3) is "yes" for symmetric monoidal categories. Let $f:X\otimes X\to X$ be the two-sided inverse of $id_x\otimes i$. As it turns out, $f$ is also the left inverse of $i\otimes id_X$. This is proven in the image linked below.
Let $\phi:X\to X^\vee$ be the composition 
$X\to X\otimes I\to X\otimes X\otimes X^\vee\to X\otimes X^\vee\to I\otimes X\otimes X^\vee\to X^\vee\otimes X\otimes X\otimes X^\vee\to X^\vee\otimes X\otimes X^\vee\to X^\vee\otimes I\to X^\vee,$
and let $\psi:X^\vee\to X$ be the composition $X^\vee\to I\otimes X^\vee\to X\otimes X^\vee\to I\to X$. These two maps are mutually inverse.
Here is a string diagram "proof" of both of the above facts.
This is the old proof with an unjustified step at the end.
A: Ok, I have no idea how to fix the invariant of string diagrams, but I have an explicit counterexample. Consider the 2-category of distributors $Dist$: its objects are small categories, its morphisms from $C$ to $D$ are bifunctors $C\times D^{op} \to Set$, its 2-morphisms are natural transformations and the composition is given by tensor product of bifunctors. It is a symmetric monoidal 2-category with monoidal product given by $\times$. Its equivalences are Morita equivalences of categories, which are the same as colimit-preserving equivalences of their presheaf categories. Any category is a dualizable object in $Dist$, the dual is given by the opposite category and the evaluation-coevaluation morphisms are given by hom-functors. We can extract a symmetric monoidal 1-category from it by discarding all noninvertible 2-morphisms and collapsing all invertible ones, i.e. it is the 1-truncation of the 1-categorical core. A simpler and similar example would be the 2-category of associative algebras with bimodule categories for morphisms, but I can't produce a counterexample in it.
Anyway, for any category $C$ its infinite cartesian power $C^{\mathbb N}$ is obviously idempotent. It's also a category of functors from $\mathbb N$, so all its morphisms and limits/colimits are pointwise. Consider now the category of nonempty finite sets $Fin_{ne}$. I claim that $\left(Fin_{ne}\right)^{\mathbb N}$ and $\left(Fin^{op}_{ne}\right)^{\mathbb N}$ are not Morita equivalent. They are certainly not equivalent as categories since $Fin_{ne}$ (and thus $\left(Fin_{ne}\right)^{\mathbb N}$) has a final object and no initial object while for $Fin^{op}_{ne}$ it is the other way round. But $Fin_{ne}$ (and thus $Fin^{op}_{ne}$) are Cauchy complete (all idempotents split), so the same is true for $\left(Fin_{ne}\right)^{\mathbb N}$ and $\left(Fin^{op}_{ne}\right)^{\mathbb N}$. Any Morita equivalence between Cauchy complete categories is realised as an actual equivalence of categories, by Theorem 2.7.3, QED.
A: From Anton Fetisov's work I think we can extract a simple example. In fact this example should be universal for the class of examples he constructs.
Recall that a connected second-countable one-manifold with boundary is a closed interval, an open interval, a half-open interval, or a circle. 
Consider the category whose objects are pairs of countable sets $(A^+,A^-)$ and where a morphism $(A^+, A^-) \to (B^+, B^-)$ is a second-countable 1-manifold with boundary, whose boundary equals $A^+ \cup A^- \cup B^+ \cup B^-$, and such that each closed interval component has one boundary point in $A^+ \cup B^-$ and one in $A^- \cup B^+$ (up to isomorphism).
Composition of a morphism $(A^+ , A^-) \to (B^+, B^-)$ with a morphism $(B^+,B^-) \to (C^+,C^-)$ is obtained by gluing the manifolds along $B$.
A symmetric monoidal structure is given by disjoint union of sets and morphisms. The identity is a union of closed intervals connecting each element to itself.
If I did this with finite sets and compact manifolds, so that the only components were closed intervals and circles, I would have constructed the free symmetric monoidal category on one generator and its dual. Because I allow infinite sets, I must necessarily allow connected components that are not closed.
The element $(\mathbb N,\emptyset)$ is idempotent as $(\mathbb N,\emptyset) \otimes (\mathbb N,\emptyset)= (\mathbb N \cup \mathbb N,\emptyset \cup \emptyset) = (\mathbb N , \emptyset)$.
It has dual $(\emptyset, \mathbb N)$, with the unit and counit each given by a union of closed intervals connecting the two copies of $\mathbb N$, where the relations hold by an infinite disjoint union of the usual stringy proof.
However, it is not self dual, as $(\mathbb N,\emptyset)$ and $(\emptyset,\mathbb N)$ are not isomorphic, as every map between them has all elements connected to half-open intervals (because $B^- \cup A^+$ is empty), so the composition also has all elements connected to half-open intervals and thus isn't the identity.
A: If $C$ is not assumed to be symmetric, then the answer to questions 1 and 2 is no.  Let $p^* \dashv p_* \dashv p^!$ be a fully faithful adjoint triple with $p_* : A \to B$.  (For instance, $A= \mathrm{Top}$ and $B=\mathrm{Set}$ with $p_*$ the forgetful functor, $p^*$ the discrete topology, and $p^!$ the indiscrete one.)  Then $F = p^* p_*$ is an idempotent comonad on $A$ and $U = p^! p_*$ is an idempotent monad, and $F \dashv U$.  Thus, $F$ is well-idempotent and dualizable in the endofunctor category $C = [A,A]^{\mathrm{op}}$ with its composition monoidal structure, but is not generally self-dual.
A: Here are a few more facts which seem relevant here. Let me adopt the Boyarchenko-Drinfeld terminology from the preprint I linked to in the comments above, and use the term "closed idempotent" for what I called "well-idempotent" in the question, and "open idempoent" for "co-well-idempotent".


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*The first point is that if $X$ is a closed idempotent via $r: I \to X$ and simultaneously an open idempotent via $i: X \to I$, then there are two additional coherences one should ask for: (a) $ri = 1_X$ and (b) $1_X \otimes ir = 1_{X}$. In this case, let's say that the pair $(r,i)$ exhibits $X$ as a "clopen idempotent". I'd call (a) the "splitting condition" and (b) the "stability condition".

*But we get these coherences for free: if $X$ is exhibited as a closed idempotent by $r$ and an open idempotent by $i$, then there is an $i'$ such that the pair $(r,i')$ exhibits $X$ as a clopen idempotent. The proof is similar to the proof that every equivalence can be made into an adjoint equivalence.

*Every clopen idempotent is self-dual, with unit given by $r \otimes r$ and counit given by $i \otimes i$. This is easy to show using (a) and (b).

*Conversely, if a closed idempotent has a dual (and hence is self-dual), then it is clopen.
So this was really a question about clopen idempotents. Also, everything seems to continue to work in the braided case. Two other interesting facts proved in the Boyarchencko-Drinfeld preprint:


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*If $X$ is a closed idempotent, then the braiding $X \otimes X \to X \otimes X$ is equal to the identity.

*If $X$ is a closed idempotent, then the map $r$ exhibiting it as such is uniquely determined. (In particular, above we must have $i' = i$.)
