Re: A question on adjoints

[Michael Barr <barr@math.mcgill.ca>]

Actually, F isn't even a functor.  The unique arrow 0 --> Ub has to give a
canonical arrow F0 = 1 --> b, which there isn't.  You could choose one, of
course, but it could not be functorial.

Actually, I realized the answer to my question cannot be yes.  Here's why.
Let A be some complete category to be specified later.  Let d be a fixed
object of A.  Let B be set\op and Fa = Hom(a,d).  The right adjoint is
given by b |---> d^b.  It is not entirely trivial to show this, but if my
answer were "yes", then you could show that the class of objects that were
equalizers of powers of d would be complete.  It is obviously closed under
products but, over 40 years ago, Isbell gave an example in which it was
not closed under equalizers.

This much is true: if there is an equalizer of the form a --> UFa ===> Ub,
then a ---> UFa ===> UFUFa is an equalizer.

Michael

On Tue, 18 Mar 2008, Vaughan Pratt wrote:

> Isn't the following a counterexample?
>
> Let A = Set and let B = A\{0} (the category of nonempty sets).  Let F send
> the empty set in A to the singleton set in B, and otherwise let F and U be
> the evident identity functors between A and B.  Similarly let \eta and
> \epsilon be the identity natural transformations, except for \eta_0 which can
> only be the unique function from 0 to 1.   Naturality of \eta and \epsilon
> depends on both being the identity, except for \eta_0 but that's from the
> initial object so all its diagrams commute.
>
> Then 0 equalizes the two arrows from U1 to U2 but \eta_0 does not equalize
> UF\eta a and \eta UFa since the latter two are both 1_1 in A whence they are
> equalized by 1.
>
> Vaughan
>
> Michael Barr wrote:
>>  I guess I am getting old and dumb.  This question should have been a snap
>>  for me years ago.  It is old fashioned, only a 1-categorical question and
>>  not about internal vs. external.
>>
>>  Suppose F: A --> B is left adjoint to U: B --> A.  Suppose a is an object
>>  of A and b, b' objects of B such that there is an equalizer
>>    a ---> Ub ===> Ub'.  (The two arrows Ub to UB' are not assumed to be U
>>  of arrows from B.)  Does it follow that a ---> UFa ===> UFUFa is an
>>  equalizer?  The arrows are \eta a, UF\eta a and \eta UFa of course.
>>
>>  Michael
>>
>>
>
>