32

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^{-...


16

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 ...


11

An ordinary monoid is a set $M$ equipped with a map $\mu\colon M\times M\to M$ and an element $e\in M$ such that certain axioms are satisfied. Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible. The first step is to observe that ...


10

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


10

Suppose that your diagram didn't commute. For example, suppose that you started at the top left corner, and traveled to the top right corner using the two different paths available to you, and you got different answers. Then there wouldn't be a unique path from $(((A\otimes B)\otimes C)\otimes D$ to $A\otimes(B\otimes (C\otimes D)))$. Why would we care ...


9

"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 \...


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 ...


9

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 $f = 2\cdot \text{id}$.


9

The quote you mention does not say that monads are monoids in the usual sense; it says that they are monoids in the monoidal category of endofunctors of $X$. A monoid in a monoidal category (or monoid object, as suggested by Derek Elkins in a comment) is not a monoid; that's a red herring. In general, a monoid object is an oject of a category which is not ...


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.


8

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}...


8

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 boring:...


8

In any monoidal category $(C, \otimes, I)$, the unit object $I$ has the property that its endomorphism monoid $\text{End}(I)$ is commutative, by the Eckmann-Hilton argument. So if $C$ is a category in which every endomorphism monoid is noncommutative, then it cannot have any monoidal structure. A reasonably natural geometric example is given by the ...


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

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

If $G$ is an abelian group, let $C$ be the category whose objects are the elements of $G$ and with only identity morphisms ($C$ is a discrete category). $C$ is symmetric monoidal via the product in $G$. The category $\text{Aut}_\text{br} (C)$ is again discrete, it's objects are the automorphisms of $G$ ($\text{Aut}_\text{br} (C)$ is constructed out of $\text{...


7

Yes. An example is the category of presheaves over a monoidal category. The category of presheaves for any (small) category is cartesian closed. If that small category has a monoidal structure, then we can define a monoidal structure on the category of presheaves via Day convolution. This category is monoidally closed because $\mathbf{Set}$ is complete. ...


7

As you probably already know, a monoid object $M$ in the category of sets with Cartesian product is precisely a monoid in the usual sense: the unit map $u: * \to M$ takes the single-element set to the unit element of the monoid, while the multiplication $m \colon M \times M \to M$ gives the monoid multiplication. What is a comonoid object $C$ in this ...


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

A promonad 'in' category $\bf D$ is the categorical correspondent of an algebra over a ring $R$. I prefer to look at profunctors $\ F:{\bf A}^{op}\times{\bf B}\to\bf{Set}\ $ as their collages: considering the elements of each $F(A,B)$ as 'outer arrows' (heteromorphisms) from $A$ to $B$, thus giving a bigger category which contains (disjointly) $\bf A$ and $\...


6

The only dualisable object in a cartesian monoidal category is the terminal object. Thus, a cartesian compact closed category must be trivial. Indeed, suppose $A$ has a dual $A^*$. Since the unit object is terminal, the counit $\epsilon : A \times A^* \to 1$ is forced. Consider the unit $\eta : 1 \to A^* \times A$. This decomposes into components as a ...


6

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 $\mathsf{C}$ ...


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

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 ...


6

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 ...


6

The Barr-Beck part of the proof is the following. Theorem: Let $C$ be an essentially small abelian $k$-linear category and let $\omega : C \to \text{Vect}_f$ be an exact faithful functor from $C$ to the category of finite-dimensional $k$-vector spaces. Then this functor exhibits $C$ as the category of finite-dimensional comodules over a coalgebra $A$. ...


6

A monoid object in the category of monoids is a monoid $M$ (with multiplication $m$ and identity $e$) equipped with monoid homomorphisms $m':M\times M\to M$ and $e':\{*\}\to M$ satisfying the axioms $$m'(m'(-,-),-)=m'(-,m'(-,-))$$ and $$m'(e',-)=\mathrm{id}_M(-)=m'(-,e').$$ Since $e'$ is a homomorphism its image must be the identity. So those axioms just ...


6

$k\otimes V$ is a certain quotient of the vector space freely generated by $k\times V$, so its elements are equivalence classes of formal sums of pairs $(x,v),x\in k,v\in V$. There's no way such a thing is equal to an element of $V$, so $k\otimes V$ isn't equal to $V$.


6

This coend actually does exist: it is just $0$. Morally, the idea is an Eilenberg swindle. Suppose we had a trace operation $tr$ defined for all vector spaces, and let $A$ be any endomorphism of a vector space. Then if you take an infinite direct sum $B$ of copies of $A$, $tr(B)$ should be an infinite sum of copies of $tr(A)$. But if you add one more ...


Only top voted, non community-wiki answers of a minimum length are eligible