# Delexing a finitely complete category

Given a finitely complete category $$C$$, is there a way to obtain a "delexing" $$\operatorname{Delex}(C)$$ of $$C$$ so that the following property holds; $$\operatorname{Lex}(C,D) \cong [\operatorname{Delex}(C),D]$$ for all finitely-complete (for simplicity) $$D$$? Where on the left we have the category of left exact functors and natural transformations between $$C$$ and $$D$$ and on the right the usual functor category.

The example I have in mind here is when $$C$$ is the finitely complete category obtained from the finite limit theory of categories. If it's not possible in general, is there a subset of finitely complete categories for which it is possible?

$$\DeclareMathOperator\Lex{Lex}\DeclareMathOperator\Delex{Delex}\newcommand\Set{\mathrm{Set}}$$If you take $$D$$ to be the category of sets, you get that $$\Lex(C,\Set)$$ is the category of functor $$[\Delex(C),\Set]$$ so it is a presheaf category.
Now, any finitely presentable category $$A$$ can be obtained as $$\Lex(C,\Set)$$ (take $$C$$ to be the opposite of the category of finitely presented objects of $$A$$), and not all finitely presentable categories are presheaf categories.
So this isn't possible in general. In particular, for the example you have in mind, it is impossible as in this case $$\Lex(C,\Set)$$ is the category of categories, which is not a presheaf category (it is not a regular category or a topos for example).
In fact, it can be shown that this is possible if and only if $$\Lex(C,\Set)$$ is a presheaf category $$[I,\Set]$$, in which case you can take $$\Delex(C) = I$$ (note that in this case the category $$I$$ is unique up to Cauchy completion).