In section III.1 of P.M. Cohn's Universal Algebra a notion of *universal functor* ${\cal L} \rightarrow {\cal K}$ is defined for a representation of one category in another given by a (covariant) functor $F: {\cal L}^{\mbox{opp}} \times {\cal K} \rightarrow \mbox{Set}$. The objects part of the functor is specified by a universal property and Cohn states and proves a proposition (1.1) stating that if this exists, then it is indeed the objects part of a functor $U$ and that a certain mapping $\rho$ is a natural transformation from $I$ to $U$.

Cohn writes $\rho$ as $\rho(A) : A \rightarrow U(A)$ where $A \in \mbox{Ob} {\cal L}$. But as $A$ and $U(A)$ are objects in different categories, even given his convention of omitting certain "obvious" functors, I can make no sense of this in the general case, where the representation is not given by a forgetful functor. (To make things "type correct", $\rho(A)$ has to be an element of the set $F(A, U(A))$).

I can see how to define the morphisms part of the functor $U$ using the universal property and how to prove that $U$ is then a functor by a diagram chase in $\mbox{Set}$, but this is nothing like Cohn's proof which appears to compose morphisms from two different categories. I can also see how Cohn's proof would say something in the case where the representation is derived from a forgetful functor, but I can't even see how to formulate the claim that $\rho$ is a natural transformation from $I$ to $U$ in the general case. I doubt very much that Cohn is wrong, so what am I missing?