What is a simple example of a representable functor that does not admit an (left) adjoint?
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$\begingroup$ The left adjoint to $\hom(A,-)$ has to send a set $S$ to the coproduct of $S$ copies of $A$. This question would be more suitable on math.SE though $\endgroup$– მამუკა ჯიბლაძეCommented Dec 4 at 17:18
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1$\begingroup$ I'd say that there's even more obvious obstruction. Left adjoint preserves colimits, so it should send colimit of the empty diagram — the empty set — to the initial object of a category (or terminal object, depending on whether you take covariant or contravariant representable). So, for any category with two arrows and one object (for both of them, actually) the only Hom-functor you have is non-representable. $\endgroup$– Denis TCommented Dec 4 at 17:28
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$\begingroup$ What does "simplest" mean? $\endgroup$– John PalmieriCommented Dec 4 at 18:00
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A left adjoint must preserve initial objects, and Set has an initial object, so a very simple example of a representable functor that does not admit a left adjoint is any representable functor on a category without an initial object. For example, the discrete category on two objects, or the free category on two parallel arrows, or such things.
(This of course is not exhaustive of the representable functors which do not admit left adjoints. In general, these are the functors represented by objects which do not admit all set-sized copowers, with an initial object as the particular special case of a 0-ary copower.)
answered Dec 4 at 17:28
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$\begingroup$ I see now that at the same time as I was writing this up, Denis T made an essentially identical comment. $\endgroup$ Commented Dec 4 at 17:29