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

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