1
$\begingroup$

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?

$\endgroup$

1 Answer 1

11
$\begingroup$

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).

$\endgroup$
12
  • 2
    $\begingroup$ However, if $C$ is empty, then $C \times A \cong C$. $\endgroup$
    – Zhen Lin
    Aug 9, 2014 at 17:54
  • 2
    $\begingroup$ Oh, you're right! In fact, the empty category is obviously an example; I'll edit that in. $\endgroup$ Aug 9, 2014 at 17:55
  • $\begingroup$ 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. $\endgroup$ Aug 9, 2014 at 18:12
  • $\begingroup$ @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 $A$. $\endgroup$ Aug 9, 2014 at 18:22
  • 1
    $\begingroup$ How do you find such an $A$ if $C$ is large? $\endgroup$ Aug 9, 2014 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.