# Isomorphic morphisms. A 27-morphism category

Two morphisms of category $$\ \mathbf C\$$ are isomorphic to one another $$\ \Leftarrow:\Rightarrow\$$ they are the opposite edges that are drawn horizontally (aimed East) of a commutative square that has the vertical edges (aimed North) being isomorphisms of $$\ \mathbf C$$.

Problem What is the minimum total number of morphisms of a category such that there are isomorphic morphisms $$\ f\$$ and $$\ u,\$$ and another isomorphic pair $$\ g\$$ and $$\ v,\$$ and the compositions $$\ g\circ f\$$ and $$\ v\circ u\$$ exist but are not isomorphic?

I have an example of a category, as described above, that has a total number of $$27$$ morphisms. (No, I've touched NO computer :) ).

Now, that @HenrikRüping has provided his excellent example (most likely minimal), let me mention that my example was a monoid too (but of course) of all maps into itself of a 3-element set.

• It feels good to have more than one example. Sep 28 at 22:55
• @LSpice, thank you for your gentle and cultural editing. Sep 29 at 0:36
• More mathematically, I'm not sure I understand your example. Which elements of your monoid are invertible? Namely, what are the north-pointing maps in your commutative square? Sep 29 at 5:05
• @DavidRoberts and Henrik, I fooled myself (confusing in this case an automorphism of the whole monoid and of morphism-isomorphisms). Sorry. Sep 29 at 5:12
• @WlodAA no worries, glad to have helped sort this out. We can delete all our earlier discussion now, if you like. Sep 29 at 5:13

I see one example with 7 morphisms. It is a subcategory of the category of groups. The only object is the the Klein 4-group $$(\mathbb{Z}/2)^2$$, and the morphisms are generated by the two projections and the flip. That monoid has 7 elements, and the two projections are conjugate, and hence they are isomorphic. However $$pr_1\circ pr_1=pr_1$$ is not isomorphic to $$pr_1\circ pr_2=0$$.
Edit: I miscounted the number of elements. Originally I thought of it as a semidirect product (which it is not). However in terms of matrices, it should consist exactly of the elements $$\pmatrix{1&0\\0&1}$$,$$\pmatrix{1&0\\0&0}$$,$$\pmatrix{0&0\\0&1}$$,$$\pmatrix{0&0\\0&0}$$,$$\pmatrix{0&1\\1&0}$$,$$\pmatrix{0&0\\1&0}$$,$$\pmatrix{0&1\\0&0}$$.
• Note that all monoids of cardinality $\leq 10$ have already been enumerated. So I think the enumeration of all categories with at most 6 morphisms should also be doable. So if people really want to know the smallest one, then brute force could be the solution in this case.