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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

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

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

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 $B_n$.)


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

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 ...


4

So to clarify, this is true in a spherical category (when you assume that left and right traces are equal, this is the definition of a spherical category). The best way (IMO) to prove this is through graphical calculus. Here is a picture (which I hope is not too horrible). The only non-obvious equality is the first: it follows from the monoidality of the ...


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

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

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)) = ...


4

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 ...


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

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, ...


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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