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.

  • $\begingroup$ It feels good to have more than one example. $\endgroup$
    – Wlod AA
    Sep 28, 2021 at 22:55
  • 1
    $\begingroup$ @LSpice, thank you for your gentle and cultural editing. $\endgroup$
    – Wlod AA
    Sep 29, 2021 at 0:36
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    $\begingroup$ 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? $\endgroup$
    – David Roberts
    Sep 29, 2021 at 5:05
  • $\begingroup$ @DavidRoberts and Henrik, I fooled myself (confusing in this case an automorphism of the whole monoid and of morphism-isomorphisms). Sorry. $\endgroup$
    – Wlod AA
    Sep 29, 2021 at 5:12
  • $\begingroup$ @WlodAA no worries, glad to have helped sort this out. We can delete all our earlier discussion now, if you like. $\endgroup$
    – David Roberts
    Sep 29, 2021 at 5:13

1 Answer 1


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 $\pmatrix{1&0\\0&1}$,$\pmatrix{1&0\\0&0}$,$\pmatrix{0&0\\0&1}$,$\pmatrix{0&0\\0&0}$,$\pmatrix{0&1\\1&0}$,$\pmatrix{0&0\\1&0}$,$\pmatrix{0&1\\0&0}$.

  • $\begingroup$ Excellent, very good! Thank you. Most likely, your example is minimal. But let me wait for a while, just in case, for an ultimate conclusion. $\endgroup$
    – Wlod AA
    Sep 28, 2021 at 6:07
  • $\begingroup$ 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. $\endgroup$
    – M.D.
    Sep 29, 2021 at 10:32

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