Reuben wrote approximately:

> Concretely (but with mistakes I need help with) note that \\((f \otimes g) \circ (f' \otimes g') = (f \circ f') \otimes (g \circ g') \\) which is to say that \\( \otimes \\) is a functor as Anindya noted (but I suppose this also needs to be shown).

What's "this"?

You _don't_ need to show \\(\otimes\\) is a functor; we're assuming we have a monoidal category \\(\mathcal{C}\\) so we're assuming among other things that \\(\otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}\\) is a functor?

You _do_ need to show that \\((f \otimes g) \circ (f' \otimes g') = (f \circ f') \otimes (g \circ g') \\); that's Puzzle 278.

More precisely, _you_ don't need to show it, but _someone_ needs to show it.

> I've described this as an equality, but is that too strong?

I suggest taking [the definition of functor](https://forum.azimuthproject.org/discussion/2213/lecture-38-chapter-3-databases/p1), applying it to \\(\otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}\\), and seeing what each clause gives, concretely. There is nothing so enlightening, nor soothing to the soul, as taking a definition and seeing what it says in a specific case.