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2

You ask: wouldn't it make the most sense to just use the monoidal product from the "mother category"? but this question does not make sense, really. To judge if something makes more sense or less sense than something else, you have to spell out what you are trying to achive. Sometimes, the direct sum is the correct operation to turn modules into a ...


2

Every object in an additive category is a group and a cogroup, canonically, with respect to the monoidal structure of direct sum. This reflects the fact that vector spaces, and more generally, modules, are actually groups. So there's nothing there. Hopf algebras weren't invented to cause pain, much less because of the desire to decorate the monoidal ...


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The direct sum of $k$-vector spaces makes $\mathbf{Vec}_k$ into a cocartesian monoidal category, not a cartesian one.


1

This is indeed not true. As you observe, $T(V\oplus W)$ is the coproduct of $T(V)$ and $T(W)$ in the category of $K$-algebras, which is typically larger than $T(V)\otimes_K T(W)$. For a very simple example, take $V=W=K$. Then $T(V)$ and $T(W)$ are both polynomial rings in one variable over $K$; write $T(V)=K[x]$ and $T(W)=K[y]$. Then $T(V)\otimes_K T(W)\...


2

Let $A$ be an associative algebra. You probably know about the center $$Z(A) = \{ a \in A \mid \forall b \in A, ab = ba \},$$ the set of elements that commute with all the others. It's easy to show that this is a commutative subalgebra of $A$. It's a rather interesting construction to study. Now consider a monoidal category $\mathcal{C}$, morally an "...


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It's plain to me that they can exist, but ... are there any useful examples? Absolutely. Perhaps the historical motivating example is categories of representations of quantum groups, which can be used to build knot and link invariants such as the Jones polynomial. If you try to play this sort of game with a symmetric monoidal category it becomes very ...


5

It is not that you want to work with an non-symmetric braiding. Sometimes you have to work with one because the category you have in hand is simply a non-symmetric braided category. The canonical example is the category of braids. You want to study braids, and they are a braided monoidal category, but alas it is not a symmetric category. There are many ...


2

When working with 2-categories, the notion of completeness is not quite as clear-cut as in 1-categories. See the nlab for more information. It should be relatively straightforward to check directly that monoidal categories have all 2-limits, though, and that the forgetful functor to $\mathsf{Cat}$ preserves them. For instance, the product of some monoidal ...


4

From either definition of adjunction it follows that $F ⊣ G$ iff $G^{\mathrm{op}} ⊣ F^{\mathrm{op}}$. For a monoidal category to be monoidal closed it is enough that every $- ⊗ B$ be left adjoint. The dual of this is that $- ⊗ B$ is right adjoint. So no, the dual of a monoidal closed category is not itself closed (instead it is "coclosed", although this is ...


3

According to this MO question, and more precisely this answer by Jacob Lurie, the answer is in general no. To sum up the answer, if you have a monoidal functor $F : \mathcal{A} \to \mathsf{Mod}_R$, then $F(I) \cong R$, and thus when $F$ if fully faithful you can recover $R$ as $\operatorname{End}(I)$. This then implies that $F$ is given by $F(A) = \...



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