The following is a completement to the answer by Yonatan. I managed to work it out after I read what he had done.

It is basically about proving the converse to the implication he showed in his answer: If I'm correct one can prove that for a locally presentable category $A$, the following are equivalent:

$(1)$ $A$ is dualizable in $Pres^L$

$(2)$ $A$ is a retract of a presheaf category in $Pres^L$

$(3)$ $A$ can be constructed from an ideal $C_0 \subset C$ as described in Yonatan answer.

This being said, I'm not excluding that a better result can be obtained (for example a more canonical presentation of dualizable objects), though the example given by Yonatan seem to suggest that this is already quite close to be optimal.

$(3) \Rightarrow (2) \Rightarrow (1) $ is discussed in Yonatan answer, I'll prove $(1) \Rightarrow (2)$ and $(2) \Rightarrow (3)$ separately.

For $(1) \Rightarrow (2)$. Let $A$ be a dualizable object in $Pres^L$, its dual $A^*$ is isomorphic to the category of left adjoint functors $[A,Set]$. The fact that $A$ is dualizable means that there is a coevaluation map $Set \rightarrow A \otimes A^*$, which is fully described by the image of the singleton, giving a specific object $c \in A \otimes A^*$.

As any object of $A \otimes A^*$, $c$ can be written as a colimits of pure tensor:

$$ c = colim_{i \in I} a_i \otimes \chi_i $$

where $a_i \in A$ and $\chi_i \in A^* = [A,Set]$. The evaluation-coevaluation relation translate into:

$$ colim_{i \in I} \chi_i (u) \times a_i \simeq u $$

functorially in $ u \in A$ (where $\times$ above denote the cotensoring objects by sets, i.e. the coproduct of several copies of the same object).

It means that the $\chi_i$ together form a cocontinuous functor $A \rightarrow Set^I$, And $F \mapsto colim F(i) \times a_i$ form a cocontinuous functor from $Set^I$ to $A$, whose composite $A \rightarrow Set^I \rightarrow A$ is naturally isomorphic to the identity, hence $A$ is a retract of $Set^I$.

I now give a sketch of proof for $(2) \Rightarrow (3)$:

Assume that one has two morphisms in $Pres^L$, $f:A \rightarrow \widehat{C}$ and $g:\widehat{C} \rightarrow A$ with $g \circ f \simeq_{\Theta} Id$.

Then $P = f \circ g : \widehat{C} \rightarrow \widehat{C}$, is idempotent $P^2 \simeq_{f \theta g} P $, and moreover the isomorphism is an associative multiplication $P^2 \rightarrow P$ (I'm goging to say that $P$ is a non-unital monad), and the category $A$ identifies with the algebras for this multiplication (in the non unital sense), whose structure maps is are isomorphisms $PX \overset{\sim}{\rightarrow} X$.

Using that $P$ commutes to coproduct, this makes $P \coprod Id$ into a cocontinuous (unital) monad on $\widehat{C}$, whose algebras are the $P$-algebra (as a non-unital monad), and hence one has a category $C_P$, whose objects are these of $C$ and morphism from $c$ to $c'$ are morphisms from $c$ to $(Id \coprod P) c'$, i.e. either morphism from $c$ to $c'$, or morphism from $c$ to $P c'$ (composition of the morphism of the second type involving the use of the multiplication maps $P^2 \rightarrow P$

The map of the form $c \rightarrow P c'$ form an ideal in $C_P$, and because of the relation $P^2 \simeq P$ they satisfies the additional condition in Yonatan answer.

Now presheaves over $C_P$ are the same as presheaves over $C$ with an algebra structure for the monad $Id \coprod P$, i.e. algebra for the non-unital monade $P$. One can then check that the additional condition in Yonatan answer (with respect to the ideal mentioned above) single out the $P$-algebra for which the structure map is an isomorphism, and hence it corresponds exactly to the category $A$ we started from.

stablelocally presentable category. $\endgroup$ – Simon Henry Jul 3 '19 at 20:35