* Re: A question on adjoints
@ 2008-03-19 23:41 Vaughan Pratt
0 siblings, 0 replies; 4+ messages in thread
From: Vaughan Pratt @ 2008-03-19 23:41 UTC (permalink / raw)
To: Categories list
So it *is* a counterexample---to the notion that any old graph theorist
can do category theory. Focusing on the naturality, I forgot about
functoriality (done that before). More embarrassing is not thinking to
perform the easiest test of all category theory, F(0) = 0. And most
embarrassing is thinking that Mike could have overlooked such an easy
example. Sorry, Mike!
Vaughan
Michael Barr wrote:
> 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.
^ permalink raw reply [flat|nested] 4+ messages in thread
* Re: A question on adjoints
@ 2008-03-19 18:43 Michael Barr
0 siblings, 0 replies; 4+ messages in thread
From: Michael Barr @ 2008-03-19 18:43 UTC (permalink / raw)
To: Categories list
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
>>
>>
>
>
^ permalink raw reply [flat|nested] 4+ messages in thread
* Re: A question on adjoints
@ 2008-03-19 5:43 Vaughan Pratt
0 siblings, 0 replies; 4+ messages in thread
From: Vaughan Pratt @ 2008-03-19 5:43 UTC (permalink / raw)
To: Categories list
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
>
>
^ permalink raw reply [flat|nested] 4+ messages in thread
* A question on adjoints
@ 2008-03-18 18:11 Michael Barr
0 siblings, 0 replies; 4+ messages in thread
From: Michael Barr @ 2008-03-18 18:11 UTC (permalink / raw)
To: Categories list
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
^ permalink raw reply [flat|nested] 4+ messages in thread
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