Adjoint functor theorem finite version

Question

Take a pair of locally finite categories $\mathbf A,\mathbf B$ ($\mathbf C$ is locally finite if there are a finite set of morphisms $A\rightarrow B$, for every fixed pair $A,B$). Suppose $B$ is finitely complete, finitely well-powered (that is, subojects form a finite poset) and has a cogenerator, does every limit-preserving functor $R:\mathbf B\rightarrow\mathbf A$ have a left adjoint? I would say yes (and prove it basically the same way as the special adjoint functor theorem), is it correct?

Answer 1

Sure, this works fine. Let's take a quick look. As usual $R$ has a left adjoint if and only if each comma category $a/R$ has an initial object, giving the unit of the pending adjunction at $a$. Now since the forgetful functor $a/R\to \mathbf B$ is faithful and creates limits (using continuity of $R$), including monomorphisms, we have that $a/R$ is finitely complete, finitely well-powered, and locally finite. Furthermore if $b_0$ is a cogenerator of $\mathbf B$ then the set of all objects of the form $a\to R(b_0)$ (there are only finitely many, using local finitude of $\mathbf A$!) is a finite cogenerating set of $a/R$, by the same argument as usual. So $a/R$ is finitely complete, locally finite and finitely well-powered with a finite cogenerating set. Then the minimal subobject of $a\to \prod_{f:a\to R(b_0)} R(b_0)$ is initial in $a/R$, again as usual.

From a higher level, the usual SAFT is proved for categories which have hom-sets, subobject lattices, limits, and a cogenerating set (equivalently given completeness, a cogenerator), all of which are bounded in size by the cardinal of the set-theoretic universe, call it $\Omega$. Now set-theoretically the function of $\Omega$ is that it's an inaccessible cardinal. $\aleph_0$ is also an inaccessible cardinal, although it's sometimes manually ruled out from the definitions. Thus it would be very strange if this argument wouldn't work: the proof would have to depend somehow on the uncountability of the class of all sets.