We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B? Michael [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Without naturality we are just asking that Hom(A,C) and Hom(B,C) have the same cardinality, so this is not a very strong restriction. For example the category of non zero finite dimensional vector spaces over an infinite field K satisfies that for any four objects W,X,Y,Z, Hom(W,X) is in (non natural) bijection with Hom(Y,Z) simply because they are both infinite sets of the cardinality of K.
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Objects {A, B}, Arrows: f: A --> B, g: B --> A, denote s = gf, t = fg. equations: ss = s tt = t, it follows fs = tf, gt = sg. the hom sets are all of cardinality 2. did fast checking, needs careful checking to be sure. e. El 10/4/16 a las 15:21, Michael Barr escribi??:
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Michael, A simple example is given by the category with exactly two objects A and B, Hom(A,A), Hom(A,B), and Hom(B,B) are all the set of integers greater than or equal to zero, where Hom(B,A) is the set of integers greater than or equal to one, and where composition is addition. Note that 0 is the identity, but it is impossible for any composition involving an arrow B -> A to be 0. To get a finite example, work with the set (Z/n)* of non-zero integers modulo some non-prime integer n. Take Hom(A,A), Hom(A,B), and Hom(B,B) all to be (Z/n)*. Take Hom(B,A) to be the subset of (Z/n)* consisting of those elements without a multiplicative inverse. Take composition to be multiplication. This time, 1 is the identity, and it is impossible for any composition involving an arrow B -> A to be 1 (note that if m does not have a multiplicative inverse, then neither does any product lm or ml, so that composition is well-defined). Best wishes, Richard On Tue, Oct 04, 2016 at 02:21:43PM -0400, Michael Barr wrote:
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Michael, Consider (Z×Z×Z)-modules A = Z×Z×1, B = Z×1×Z where Z denotes the ring of integers. The full subcategory {A, B} is a (counter)example. Indeed, all the homsets Hom(A,A)=Z×Z×1, Hom(A,B)=Z×1×1, Hom(B,A)=Z×1×1, and Hom(B,B)=Z×1×Z are countable. There is no pair (f: A->B, g: B->A) s.t. g f = id. -- Toshiki Kataoka 2016-10-05 3:21 GMT+09:00 Michael Barr <barr@math.mcgill.ca>:
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Michael Barr hat am 04.10.16 um 14:21 Uhr geschrieben:
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Yes, you can. Take a category with your two objects A and B, Hom(X,Y) the set of natural numbers, composition being addition. Now remove 0 from Hom(A,B). Best regards Thorsten [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
PPS - Let me try again, attempting to remember all the conditions this time! (In the categories I described in the PS, the two objects are in fact isomorphic.) Take Hom(A,A), Hom(B,B), and Hom(A,B) to be the integers 0 and 1, take Hom(B,A) to be the integers 0 and 2. Take composition to be multiplication modulo 2. Since multiplication with 0 or 2 is always zero modulo 2, it is obvious that composition is well-defined, and that no composition involving an arrow B -> A can be 1 (the identity), so that A and B are not isomorphic. I think this example must be minimal. On Wed, Oct 05, 2016 at 10:27:43PM +0200, Richard Williamson wrote:
PS - Sorry, the finite example needs slightly tweaking. We can take both Hom(A,A) and Hom(B,A) to consist of those elements without a multiplicative inverse, along with 1. Otherwise, the example is the same as I described. It is still the case that composition is well-defined: multiplying by 1 does not affect the reason for this.
We could also take all of the four Hom sets to be the ones I just described. I think that, in this case, for n=4, we might have the minimal example for categories with two objects; can anybody find a smaller one?!
On Wed, Oct 05, 2016 at 10:00:33PM +0200, Richard Williamson wrote:
Dear Michael,
A simple example is given by the category with exactly two objects A and B, Hom(A,A), Hom(A,B), and Hom(B,B) are all the set of integers greater than or equal to zero, where Hom(B,A) is the set of integers greater than or equal to one, and where composition is addition. Note that 0 is the identity, but it is impossible for any composition involving an arrow B -> A to be 0.
To get a finite example, work with the set (Z/n)* of non-zero integers modulo some non-prime integer n. Take Hom(A,A), Hom(A,B), and Hom(B,B) all to be (Z/n)*. Take Hom(B,A) to be the subset of (Z/n)* consisting of those elements without a multiplicative inverse. Take composition to be multiplication. This time, 1 is the identity, and it is impossible for any composition involving an arrow B -> A to be 1 (note that if m does not have a multiplicative inverse, then neither does any product lm or ml, so that composition is well-defined).
Best wishes, Richard
On Tue, Oct 04, 2016 at 02:21:43PM -0400, Michael Barr wrote:
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
How about two objects A and B, with hom(A,A)=hom(B,B) the natural numbers, and hom(A,B)=hom(B,A) the positive integers, and all compositions given by addition? Steve.
On 5 Oct 2016, at 5:21 AM, Michael Barr <barr@math.mcgill.ca> wrote:
We all know that if Hom(A,-) is naturally equivalent to Hom(B,-), then A is isomorphic to B. But can you find a category in which for each object C, Hom(A,C) is isomorphic to Hom(B,C) but no naturality of the isomorphism without A being isomorphic to B?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
participants (7)
-
Eduardo Julio Dubuc -
henry＠phare.normalesup.org -
Michael Barr -
Richard Williamson -
Steve Lack -
Thorsten Palm -
Toshiki Kataoka