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Does there exist a category C which such that there is no functor $F:C \rightarrow D$ with $D\not\cong C$ which has a left (or right) adjoint?

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up vote 10 down vote accepted

The empty category trivially satisfies this (there are no functors at all from a nonempty category to the empty category), but no other such category exists. Let $A$ be any category with a terminal object $1$, and consider the projection $C\times A \to C$. This has a right adjoint $C\to C\times A$ given by $c\mapsto (c,1)$. However, these functors are not equivalences unless $C$ is empty or $A$ is equivalent to the terminal category (and more generally if $C$ is small (or even accessible) it is easy to find an $A$ such that $C\times A$ cannot be equivalent to $C$ by any functor).

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However, if $C$ is empty, then $C \times A \cong C$. – Zhen Lin Aug 9 '14 at 17:54
Oh, you're right! In fact, the empty category is obviously an example; I'll edit that in. – Eric Wofsey Aug 9 '14 at 17:55
Here I am assuming that by "$D\not\cong C$" you really mean that the adjunction in question is not an equivalence, which seems to me like the most natural question to ask. If you really mean to ask whether there exists any equivalence between $C$ and $D$ and want to ask about large categories with no restrictions whatsoever, the question seems like it might be rather delicate set-theoretically. – Eric Wofsey Aug 9 '14 at 18:12
@EricWofsey, I don't think there are any reasonable set-theoretic issues. If you recall that (any) set theory is closed under set-theoretic operations, then it'll be obvious, that due to cardinality reasons, you may always find such $A$. – Michal R. Przybylek Aug 9 '14 at 18:22
How do you find such an $A$ if $C$ is large? – Eric Wofsey Aug 9 '14 at 18:23

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