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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 and shown by Adámek and Rosický, 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.

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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.

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    $\begingroup$ A belated welcome to MO, Professor Rosický. Looking forward to reading more from you. $\endgroup$
    – Todd Trimble
    Jul 12, 2015 at 9:15
  • $\begingroup$ Ah, such a simple example! Thank you! $\endgroup$
    – Tim Campion
    Jul 13, 2015 at 3:02

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