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Suppose we have categories $\mathcal{C}$ and $\mathcal{D}$ with some appropriate adjectives like local smallness along with a pair of functors $F: \mathcal{C} \rightarrow \mathcal{D}$ and $G: \mathcal{D} \rightarrow \mathcal{C}$. Suppose further that for every pair of objects $A$ in $Ob(\mathcal{C})$ and $B \in Ob(\mathcal{D})$ we have a bijection of sets $$ \text{Hom}_{\mathcal{D}}(F(A), B) \simeq \text{Hom}_{\mathcal{C}}(A, G(B)), $$ but say we do not assume a priori that this is an isomorphism of bifunctors, that is, presumably naturality in $A$ or $B$ may fail. Is this possible? I strongly suspect it is, because naturality seems like an essential part of an adjunction, but are there explicit counter examples? Or will all such counter examples be fairly artificial constructions like a finite category, etc?

As an extension, suppose we knew (via some adjoint functor theorem) that (WLOG) $F$ did indeed have a right adjoint. Would the above bijection be enough to conclude that it must be $G$, or can there be non-isomophic functors that agree with an adjoint pair on objects only?

Similarly, if we were able to show naturality in (WLOG) $A$, and we knew an adjoint existed, is there some Yoneda argument that can be made to force naturality in $B$?

I'm sure these may not be too difficult to work out for someone with experience, but I'm still relatively new to adjoint functors so I'm not sure I'd trust my own answer even if I did manage to figure anything out.

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    $\begingroup$ For the second question, if we knew the bijection was natural in $A$, then there would be a unique way (up to isomorphism) to extend the object part of $G$ to a functor such that the bijection was natural in $B$ too. That is to say, we can conclude that a functor with the same action on objects as $G$ is the right adjoint of $F$, but that functor may not be $G$. It is certainly possible to have functors with the same action on arrows and non-isomorphic actions on arrows, e.g. the identity functor and the functor sending all arrows to identity arrows on any category with only endomorphisms. $\endgroup$ – Derek Elkins Mar 5 '18 at 5:32
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For the first question, let $F$ be the duality functor on finite-dimensional vector spaces and $G$ the identity functor. One knows $F(V)\cong V$ non-naturally, so $\mathbf{Vect}(F(V),W)\cong \mathbf{Vect}(V,W)$, non-naturally. I assume this counts as a natural example, since it's the standard example of an unnatural isomorphism. Derek answered the last questions in the comments.

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  • $\begingroup$ This is exactly the kind of simple answer I was hoping for (or rather hoping couldn't happen). It seems obvious in retrospect but I had never come across it in examples. Thanks! $\endgroup$ – Luke Mar 5 '18 at 10:23
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You can also find some counterexamples when $\mathcal{C}$ and $\mathcal{D}$ only have one object each (which I'll denote $*$), so they are actually monoids (which I'll denote $C$ and $D$) and functors between them are just monoid homomorphisms. Then an adjunction $F\dashv G$ means that there are elements $c,d$ in the monoids $C,D$ such that $$dFG(x)=xd\quad \text{and} \quad GF(y)c=cy$$ for all $x\in D$ and $y\in C$. If $C,D$ are groups, then this means that the composites $FG$ and $GF$ are conjugated to the identity, hence they must be isomorphisms, and thus $G$ and $F$ are isomorphisms themselves. On the other hand, the existence of a bijection $\operatorname{Hom}_\mathcal{D}(F(*),*)\cong \operatorname{Hom}_\mathcal{C}(*,G(*))$ only means that the underlying sets of the two monoids or groups are in bijection, which can happen even when the monoids or groups are not isomorphic.

Suppose now that for example $F$ is an isomorphism; then its inverse $F^{-1}$ is also an adjoint (both right and left), but the bijection $\operatorname{Hom}_\mathcal{D}(F(*),*)\cong \operatorname{Hom}_\mathcal{C}(*,G(*))$ is independent of $G$, so it can exist even when $G$ is not the adjoint of $F$. In fact, this bijection is given by the inverse of $F$, so it is natural in the first argument whatever $G$ is, but it is not necessarily natural in the second.

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