Let $C$ be a symmetric monoidal category.


* 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:**


1. If $X$ is idempotent and dualizable, then is $X$ self-dual?
2. If $X$ is well-idempotent and dualizable, then is $X$ self-dual?
3. 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.



If a $\otimes$-idempotent object has a dual, must it be self-dual?
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.](https://i.stack.imgur.com/FDNYK.jpg)


[This is the old proof with an unjustified step at the end.](https://i.stack.imgur.com/HWgBQ.jpg)


