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16

This is my understanding of this yoga. It may not be exactly what you seek and may differ from another person's point of view. Also I apologize for my bad english. For Grothendieck, many things should have a relative version. So instead of considering just a space $X_0$, consider a morphism $f:X\rightarrow S$ thought as a family of spaces $s\mapsto X_s:=f^{-...


9

Here's a simpler example: in ring theory, picking a basis of your ring (as a module over the ground commutative ring $k$) is extra structure. But in category theory there is sometimes a "distinguished basis" (e.g. the simple objects in an abelian category), which will pass to just a basis after decategorifying. For example, Hecke algebras have a famous ...


8

It is a one-object category enriched over $M$, see http://ncatlab.org/nlab/show/enriched+category#InMonoidCat.


8

Certainly not, for roughly the same reason that if $X,Y$ are monoids, then a bijection between the underlying sets need not respect the monoid operations. If you understand this claim, you can answer your question too.


7

The $\lambda$-structure is given by taking exterior powers. This is the main motivation I know for defining $\lambda$-rings in the first place. (You need an action of $S_n$ on an $n^{th}$ tensor power $V^{\otimes n}$ to define the exterior power, which is what being symmetric monoidal gets you; in the braided monoidal case you only get an action of $B_n$.)


7

what about $\mathbb{Z}_2$ vector spaces over a field $k$? $(k,0)$ is the unit in this category and $(0,k)\otimes (0,k) = (k,0)$.


7

In general if you don't want to start with the monoidal structure, you start with a closed category. In a closed category $\mathsf C$, you have a bifunctor $$[-,-] : \mathsf C^{op} \times \mathsf C \to \mathsf C,$$ called the internal hom, and various other data that are somewhat "dual" to the axioms of a monoidal category (I put dual in quotes because this ...


7

Day convolution is a categorification of the monoid algebra construction. There is a formal analogy between the two, but one is not a literal generalisation of the other. So to address your question 3, we should not expect to recover the usual convolution from Day convolution. Let's develop the following analogy: \begin{array}{|c|c|} \hline \textbf{monoid ...


6

What's the definition of an additive monoidal category? Is it that tensor product distributes over addition of morphisms? If so, use the fact that a functor between additive categories preserves addition of morphisms iff it preserves biproducts (see for example this blog post).


6

This is the classical situation covered by the Eckmann–Hilton argument. The point is that the multiplication on $M$ defined by $\mu$ is a homomorphism for the original multiplication $\cdot$ on $M$ and that's enough for the argument given on the Wikipedia page to apply. It seems a bit pointless to repeat it here. A number of references are given on ...


6

Your observation only means that there is no object $(X,\phi)$ in the Drinfeld center which has $X$ simple and non-isomorphic to $V_e$ ($e$ being the identity element of $G$) But let $C\subseteq G$ be a conjugacy class, and let $V_C=\bigoplus_{g\in C}V_g$. Then you should be able to find a natural isomorphism $\phi_C:V_C\otimes(\mathord-)\to(\mathord-)\...


6

"Uniqueness will fail for noncommutative monoids, so here we must use commutativity of our monoids, " This is not correct. You get a morphism in $V$ also for non-commutative monoids, but it won't be a monoid morphism. This is where commutativity is used. Uniqueness holds in general: If $h : M \otimes N \to X$ is a monoid morphism with $h \circ (M \otimes \...


6

An object $X$ of an abelian category has finite length if it has a (finite) composition series: i.e., a chain of subobjects $$0=X_0<X_1<\dots<X_{n-1}<X_n=X$$ such that $X_i/X_{i-1}$ is simple for $1\leq i\leq n$.


6

Note that taking opposite category gives rise to a $2$-endofunctor: $$(-)^{op}:\mathrm{CAT}\rightarrow \mathrm{CAT}$$ on $2$-category of all categories contained in a given universe $\mathcal{U}$. This $2$-endofunctor is covariant on functors and contraviariant on natural transfromations. Using this $2$-endofunctor you can argue as follows. If $$(\mathcal{C}...


6

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

http://arxiv.org/abs/math/0401347


5

Edit, 3/25/15: In an earlier version of this answer I claimed that group objects can be defined in any monoidal category. In fact this isn't true; one of the axioms requires access to a diagonal map, and so it's standard to only define group objects in cartesian monoidal categories. But since talking about actions only uses the monoid structure this doesn't ...


5

I hope what follows will clear up your confusion. A lax monoidal functor consists of the following data (satisfying some axioms that I won't spell out): One functor $\color{red}{F : \mathsf{C} \to \mathsf{D}}$. This means that for all objects $A \in \mathsf{C}$ you have an object $F(A) \in \mathsf{D}$, and for all morphisms $f : A \to B$ in $\mathsf{C}$ ...


5

Yes. This is a corollary of Schur-Weyl duality. You need at least the additional assumption that your symmetric monoidal category is enriched over $k$-vector spaces. In general I don't see any reason to expect that the action of $k[S_n]$ is faithful; consider, for example, the special case where we only look at $1$-dimensional vector spaces. Sometimes. The ...


5

Your guess is right, it's indeed not true in general: Any choice of an $R$-module $X$ such that $X\otimes_R X=0$ and $f\neq\text{id}: X\to X$ gives a counterexample, for example you could take $R := {\mathbb Z}$, $X := {\mathbb Q}/{\mathbb Z}$ and $f = 2\cdot \text{id}$.


5

The category of $R$-modules is monoidal closed with respect to the tensor product over $R$, but the category of $R$-algebras is not (indeed, $\otimes_R$ is just the coproduct of $R$-algebras, and so it does not distribute over coproducts, as it would have to to have a right adjoint). However, it is not true either that the category of finitely presented $R$-...


5

Note, that $X^{-1} = X^\vee$ and $ev_X$ corresponds to $\delta$. To see this, notice that by the Yoneda Lemma we have a bijection $$ \operatorname{Nat}(h_{X^{-1}},\operatorname{Hom}(-\otimes X,1)) \longrightarrow \operatorname{Hom}(X^{-1}\otimes X,1), $$ where $h_{X^{-1}}(Y):= \operatorname{Hom}(Y,X^{-1})$. So $\delta\in \operatorname{Hom}(X^{-1}\otimes X,1)$...


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 ...


4

Use the Yoneda lemma! By adjunction there is a natural isomorphism $$\mathcal{C}(X \otimes I, C) \to \mathcal{C}(X, [I, C])$$ for all objects $X$ in $\mathcal{C}$. Being an isomorphism is preserved by all functors, so the natural transformation $$\mathcal{C}(X, C) \to \mathcal{C}(X \otimes I, C)$$ induced by the right unitor $\rho_X : X \otimes I \to X$ is ...


4

I think you want to be a little more careful, as a priori there might be multiple ways of writing a given object as F(X). This isn't a really problem though because if you have some object A, you know basically how to write it as F(X): just let X = G(A)! So we just have $$G(A \bullet B) \cong G(F(G(A))\bullet F(G(B))) \cong GF(G(A) \otimes G(B)) \cong G(A) ...


4

Since you are interested in higher categories, you should see the Eckmann-HIlton argument as a special case of the use of the interchange law, or exchange law. A general formulation of this is for double categories: this is a set, or class, $C$, with two category structures $\circ_1, \circ_2$ each of which is a morphism for the other. This amounts to the ...


4

You are right that the Yoneda lemma plays the key role in the proof. By the definition of convolution: $$(F \otimes G) \otimes H \approx \int^{C, D} \left(\int^{A, B} F(A) \times G(B) \times \hom(C, A \otimes B)\right) \times H(D) \times \hom(-, C \otimes D)$$ Because products in $\mathbf{Set}$ preserve coends, the above is isomorphic to: $$\int^{A, B, C, D}...


4

The additional structure is that there is a natural further quotient of the isomorphism classes you can take where you impose the additional relation $[X] - [Y] + [Z] = 0$ for every distinguished triangle $X \to Y \to Z \to \Sigma X$. As Martin says in the comments, any commutative monoid gives a discrete braided monoidal category. (Recall that a discrete ...


4

The term "linear monoidal category" doesn't mean "linear and monoidal category", but rather "monoidal (linear category)", i.e. a (weak) monoid in the monoidal bicategory of linear categories (over some fixed base ring $R$). The monoidal product of two linear categories $C,D$ has as objects pairs of objects of $C,D$ with hom-modules $\hom((a,b),(c,d)) = \hom(...


4

A module is an abelian group. (It's more useful to think of a module as the analog of a vector space, but with the set of scalars coming from a ring instead of a field. Usually, one arrives at this notion of a module in terms of "the action of a ring on a set" where the set is a module.) A monoid is a relaxation of the definition of a group. A monoid has ...



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