Reuben wrote:

> 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.

I'm glad you say that they're clearly wrong.

The most scary thing is that you're tensoring a morphism \\(id\\) with an object \\(I\\) - or at least, that's what it looks like, since I said in the lecture that \\(I \in \mathbf{Ob}(\mathcal{C}) \\) is the unit object for the tensor product, while \\( id \\) is a standard name for an identity morphism from some object to itself - you're not saying which one. Overall, you seem to be saying the morphism \\(id\\) is equal to the object \\(I\\), which can't possibly be true.

There are people who tensor morphisms with objects, but nothing I said in the lecture enouraged doing that, or said what it would mean... so let's only tensor objects with objects, and morphisms with morphisms, okay? That's what the functor

\[ \otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C} \]

lets us do, and this should be enough for our puzzles here.

> 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.

I'm glad you say that they're clearly wrong.

The most scary thing is that you're tensoring a morphism \\(id\\) with an object \\(I\\) - or at least, that's what it looks like, since I said in the lecture that \\(I \in \mathbf{Ob}(\mathcal{C}) \\) is the unit object for the tensor product, while \\( id \\) is a standard name for an identity morphism from some object to itself - you're not saying which one. Overall, you seem to be saying the morphism \\(id\\) is equal to the object \\(I\\), which can't possibly be true.

There are people who tensor morphisms with objects, but nothing I said in the lecture enouraged doing that, or said what it would mean... so let's only tensor objects with objects, and morphisms with morphisms, okay? That's what the functor

\[ \otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C} \]

lets us do, and this should be enough for our puzzles here.