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.