280: my thought process: consider a monoid where morphisms I->I are elements and multiplication is composition. Then the exercise is asking to show that this monoid is commutative. This sounds like the Eckmann-Hilton argument.

Concretely (but with mistakes I need help with) note that \$$(f \otimes g) \circ (a \otimes b) = (f \circ a) \otimes (g \circ b) \$$ which is to say that \$$\otimes \$$ is a functor as Anindya noted (but I suppose this also needs to be shown). I've described this as an equality, but is that too strong?

Then \$$id \$$ = \$$id \circ id\$$ = \$$(I \otimes id) \circ (id \otimes I) \$$ = \$$(I \circ id) \otimes (id \circ I) \$$ = \$$I \otimes I \$$ = \$$I \$$ . Again, lots of these equalities are clearly wrong. Not least because there's no morphism from X to \$$I \otimes X \$$. Moreover, id is a morphism and I is an object. They're definitely not equal, although for id restricted to I, they might be isomorphic.

One continues as on https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument to show that a.b = b.a: I'm not bothering with that yet, because I want to debug my conceptual confusion this far / check if I'm doing something reasonable.

Edit 1: hmm okay, \$$(id \circ I) \$$ isn't even well-typed.

Edit 2: aha! maybe what I should show is, instead: for any morphism f : \$$I \to I \$$, f is isomorphic to \$$id \$$. Then commutativity of fg and gf follows trivially. This is easier: \$$f \approx (f \otimes f) = ((f \circ id) \otimes (id \circ f) ) = ((f \otimes id) \circ (id \otimes f) ) = (id \otimes id ) \approx id \$$ . This is still clearly wrong somehow, but maybe less wrong?