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14

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 ...


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

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

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 ...


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

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

"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

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 ...


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

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 ...


5

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


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

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

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 $B_n$.)


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

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

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)) = ...


4

Let $\mathcal{C}$ be the following category: The objects are pairs $(I, X, p)$ where $I$ and $X$ are sets and $p : X \to I$ is a surjection. The morphisms $(I, X, p) \to (J, Y, q)$ are maps $f : X \to Y$ such that $q \circ f = p$. (That means we must have $I \subseteq J$ to have a morphism.) Composition and identities are inherited from $\mathbf{Set}$. I ...


4

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 ...


4

I think you are pointing to two different phenomena, and I don't know which one you're more interested in. Can you clarify? In the second example you're just unwrapping the definition of a functor $BM \to \text{Set}$. You can always describe what a functor $C \to D$ looks like in a manner "internal to $D$" in a sort of trivial way: of course it's the same ...


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 ...


4

There is a chain of forgetful functors which progressively forgets the various operations in the structure: $$\mathrm{Mod_R}\to\mathrm{Ab}\to\mathrm{AbMon}\to\mathrm{Set}$$ The interesting thing is that you can go in the opposite direction too with free functors $$\mathrm{Set}\to\mathrm{AbMon}\to\mathrm{Ab}\to\mathrm{Mod_R}$$ Each forgetful functor $U$ ...


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, ...


3

The pentagon identity is a commutative diagram of $2$-morphisms, each one induced by $\alpha_M$. It looks as follows. Because I'm lazy, I will omit the associators from $\mathcal{C}$. $$\begin{array}{cc} \mu \circ ((\mu \circ (\mu \otimes\mathrm{id}_M)) \otimes \mathrm{id}_M) & \rightarrow & \mu \circ ((\mu \circ (\mathrm{id}_M \otimes \mu)) ...


3

The term you want is "distributes over," not "commutes with." (Whatever "$A$ commutes with $B$" means it should be symmetric in $A$ and $B$, which the condition you want is not.) Such monoidal categories are called distributive. Examples include any closed monoidal category with coproducts because in this case $A \otimes (-)$ has a right adjoint and hence ...


3

The question is not really precise. And you certainly don't look for an equivalent concept. Perhaps you are looking for the notion of a rig object internal to a rig category? This is an object $x$ equipped with morphisms $z : 0 \to x$ (zero) , $a : x \oplus x \to x$ (addition), $u : 1 \to x$ (unit) and $m : x \otimes x \to x$ (multiplication) such that the ...


3

[This answer is copied with very minor modifications from a preprint of mine "The Balanced Tensor Product of Module Categories" joint with Chris Douglas and Chris Schommer-Pries which will appear on the arxiv in the next few weeks.] Let $\mathcal{R} \cong \mathrm{Vec} \oplus \mathrm{Vec} \cdot X$ be the symmetric monoidal category consisting of pairs of ...


3

In the right upper part of the diagram, you can reduce $F(\mathsf{free} f) \circ \mathsf{inr}$ to $\mathsf{id} \otimes \mathsf{free} f : A \otimes A^{*} \to A \otimes G$. You end up with $(\mathsf{id} \otimes \mathsf{free} f) \circ (\mathsf{id} \otimes e)$, which is $\mathsf{id} \otimes e_G$. There you can swap the order with $f \otimes \mathsf{id}$ (a.k.a. ...



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