It's well-known that the category of models of any first-order theory $T$ form an accessible category if we take the *elementary embeddings* as morphisms. This is true in finitary first-order logic or infinitary first-order logic. This is hinted at by [Makkai and Paré](http://www.ams.org/books/conm/104/) and shown by [Adámek and Rosický](https://books.google.com/books?id=iXh6rOd7of0C), Thm 5.42. 


**But what happens when we take our morphisms to be arbitrary homomorphisms? Is the resulting category still accessible?** It turns out (Adámek and Rosický Thm 5.35, Makkai and Paré Thm 3.2.1) that a category is accessible if and only if it is equivalent to the category of models of a *basic theory* of infinitary first-order logic, with homomorphisms as morphisms. A basic theory is one which is axiomatized by a small set of sentences each of the form $\forall \vec x \, (\phi(\vec x) \to \psi(\vec x))$ where $\phi$ and $\psi$ are *positive existential*, i.e. built up from $\vee, \wedge, \exists$ and atomic formulas.


It seems too much to ask that every theory be "Morita equivalent" (in the sense of having equivalent categories of models-and-homomorphisms) to a theory of this restricted form, so perhaps the answer is no. On the other hand, every category of models-and-elementary-embeddings is accessible, and hence (surprisingly, to me) equivalent to the category of models-and-homomorphisms of another theory with this restricted form -- so maybe an affirmative answer wouldn't be so surprising after all.


I don't expect the answer to depend on whether we use finitary or infinitary logic, but if it does, that would be very interesting.



Do models-and-homomorphisms always form an accessible category?
For a general theory, one does not need to have $\lambda$-directed colimits for any $\lambda$.
The simpliest example is formed by sets and the axiom $(\exists x,y)(x\neq y)$; idempotents do not split here.


