Re: Further to my question on adjoints

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

No.  For example, in the category of topological abelian groups, Z is far
from injective.  Nonetheless, if you say that a group is Z-compact when it
is an equalizer of two maps between powers of Z, then an equalizer of two
maps between Z-compact abelian groups is again Z-compact.  The proof is
not direct.  As it happens, I am talking on this in our seminar tomorrow.

Even though the reals are not injective in hausdorff spaces, a space is
realcompact iff it is a closed subspace of a power of R, which turns out
to be equivalent to being an equalizer of two maps between powers of R
(that is a cokernel pair of such a closed inclusion has enough real-valued
functions to separate points) and it is clear that a closed subspace of a
realcompact space is again realcompact.  Same thing for N-compact.  In
fact, for every example I have looked at sufficiently closely.

Michael

On Mon, 12 May 2008, Eduardo Dubuc wrote:

> Consider the dual finitary question: In universal algebra in order to show
> that finitely presented objects  are closed under coequalizers it is
> essential  that a amorphism of finitely presented objects lift to a
> morphism  between the free. Is this the only way to prove it ? :
>
> " but when I look at examples, it has turned out to be true
> for other reasons."
>
> greetings  e.d.
>
>
>>
>> In March I asked a question on adjoints, to which I have received no
>> correct response.  Rather than ask it again, I will pose what seems to be
>> a simpler and maybe more manageable question.  Suppose C is a complete
>> category and E is an object.  Form the full subcategory of C whose objects
>> are equalizers of two arrows between powers of E.  Is that category closed
>> in C under equalizers?  (Not, to be clear, the somewhat different question
>> whether it is internally complete.)
>>
>> In that form, it seems almost impossible to believe that it is, but it is
>> surprisingly hard to find an example.  When E is injective, the result is
>> relatively easy, but when I look at examples, it has turned out to be true
>> for other reasons.  Probably there is someone out there who already knows
>> an example.
>>
>> Michael
>>
>>
>