It is known a version of Adjoint Functor Theorem for locally presentable categories, which says that a functor between such categories has a left adjoint iff it is continous (i.e. preserves all limits) and accessible (preserves $\lambda$filtered colimits for some cardinal $\lambda$) (see for this Theorem 1.66 of J. Adamek, J. Rosicky, "Locally Presentable and Accessible Categories", London Math. Soc. Lecture Notes, Cambridge, 1994). My question is then to find an example of a continous functor between locally presentable categories which is not accessible.

4$\begingroup$ Isn't this one of those things that only exists if we assume the negation of Vopenka's principle? $\endgroup$ – Harry Gindi Jul 6 '11 at 20:23

$\begingroup$ Thanks! It could be, but I don't know any reference. $\endgroup$ – George C. Modoi Jul 6 '11 at 20:32
The following example was coincidentally mentioned by André Joyal on the categories mailing list today; he attributed it to Mac Lane. For every infinite cardinal number k, let $G_k$ be a simple group of cardinality k. Define the functor ML: Group → Set to be the product of all the representable functors $\mathrm{Hom}(G_k,)$. Since no group can admit a nontrivial homomorphism from properclassmany of the $G_k$, this functor does indeed land (or can be redefined to land) in Set. Since it is a product of representables, it is continuous (and of course Group and Set are locally presentable), but it is not itself representable (hence has no left adjoint).

1$\begingroup$ Great! Actually I seen the example in the paper J. Adámek, V. Koubek and V. Trnková, "How large are left exact functors?", 2001, Theory and Applications of Categories 8 [2001], 377390, but I was dummy! $\endgroup$ – George C. Modoi Jul 13 '11 at 10:54

1$\begingroup$ It's also a nice short reason for why the category of groups is not cototal. $\endgroup$ – Todd Trimble♦ Jul 14 '11 at 16:01
Not really an answer, but here's what I know:
relationship with Vopenka's principle
the weak Vopenka's principle
Assuming the weak Vopenka's principle, every full and faithful continuous functor has a left adjoint (this is in fact equivalent to it: Theorem 6.22 in the AdamekRosicky book). So, under this assumption, you need to look for a non fully faithful continuous functor.
accessibility and the solution set condition
In general, a functor $F \colon A \to B$ between accessible categories is accessible iff the induced functor between the corresponding arrow categories: $$\overrightarrow{F} \colon \overrightarrow{A} \to \overrightarrow{B}$$ satisfies the solution set condition (as in Freyd's General Adjoint Functor Theorem).
Now, under the strong Vopenka's principle, we have the stronger
$F$ accessible iff $F$ satisfies the solution set condition
This is in Rosicky, Tholen  Accessibility and the solution set condition  JPAA
This coupled with the above, yields that the reflective subcategory is in fact locally presentable. Even more,
($A$ full reflective subcategory of a locally presentable $B$ $\Rightarrow$ $A$ locally presentable) is equivalent to the strong Vopenka's principle.
All of the examples I know of about continuous functors not satisfying the solution set condition tend to make essential use of size in the domain category (those in MacLane, for example), none of them being locally presentable.
the dual problem
I'm pretty sure you know this, but anyway: the dual problem ($F$ having a right adjoint) is straightforward: locally presentable categories are total, and for total categories $F$ has a right adjoint iff it is cocontinuous

$\begingroup$ Many thanks Eduardo! I'm just seeing the answer but it seems to have a lot to think! $\endgroup$ – George C. Modoi Jul 12 '11 at 20:54