Tag Info

Hot answers tagged

7

(In my version of Awodey this is page 78-9, not 91.) Freyd is just making a linguistic point, as made clear by what he says next: "Indeed, subobjects, as we have defined them, do not have subobjects" -- only objects have subobjects. A subobject is not an object, but rather a monomorphism (or, depending on your definition, an equivalence class of ...


5

In effect, the phrase "$(\infty, 1)$-category" is a cover term for a family of related concepts which are very closely related. Quasicategories are certain special simplicial sets. The theory has been extensively developed by Joyal and Lurie, and [Higher topos theory] covers a lot. As you note, Lurie simply calls these "$\infty$-categories". Simplicially ...


5

Let $A_i$ be the objects of a diagram and $L$ the a limit of the diagram. Pushing through your functor gives the diagram $\hom(-, A_i)$ and you want to show that $\hom(-, L)$ is it's limit. So suppose you have a contravariant functor $F$ and natural transformations $F \to \hom(-, A_i)$. I'm guessing you need to know how to define $F \to \hom(-, L)$. ...


5

Let $M$ be the set of all sequences $(x_1, x_2, x_3, \dots)$ of numbers from $[0,1]$ with the property that $f(x_{i+1}) = x_i$ for $i = 1, 2, 3, \dots$. If $x = (x_1, x_2, \dots)$ and $y = (y_1, y_2, \dots)$ belong to $M$, define the distance from $x$ to $y$ by $$d(x, y) = \sum_{i > 0} {{|x_i - y_i|}\over{2^i}}.$$$($Actually, $M$ is a certain subset of ...


4

Recall that the disjoint union of sets is usually constructed as $\coprod_i A_i = \bigcup A_i \times \{i\}$, but other constructions are also available. If $A$ is a set with a map $A \to I$ and fibers $A_i$ for $i \in I$, then $A$ is not necessarily equal to that disjoint union $\coprod_i A_i$; rather we have a canonical isomorphism $\coprod_i A_i \to A$. ...


3

The localization of a ring $A$ at a multiplicative submonoid $S$ is the initial ring under $A$ in which $S$ is sent to units. (i.e. there's a localization map $A\to S^{-1}A$ and if $S$ goes to units in $f:A\to B$ then $f$ factors uniquely through $A\to S^{-1}A$.) Diagrammatically, a unit is just an element whose action by multiplication is an isomorphism. A ...


3

The group algebra is a functor $\mathsf{Grp} \to \mathsf{Alg}_K$ which is left adjoint to the "group of units" functor $\mathsf{Alg}_K \to \mathsf{Grp}$, $A \mapsto A^\times$. It is defined for all unital $K$-algebras. No non-trivial ring has the property that its underlying semigroup is a group, because $0$ is not invertible. If $V$ is some $K$-module, ...


3

The exponential object is contravariant in the "exponent" variable: If $A \to B$ is a morphism, this induces a morphism of functors $$\hom(-,X^B)\cong \hom(- \times B,X) \to \hom(- \times A,X) \cong \hom(-,X^A) $$ and hence a morphism $X^B \to X^A$ (Yoneda Lemma). This morphism is natural in $X$ since the morphisms of functors above are natural in $X$. ...


3

A binary operation on a class $S$ is a function $S \times S \to S$. A binary partial operation is a partial function $S \times S \to S$. That is, it can be expressed as a function $D \to S$ for some subclass $D \subseteq S \times S$. For example, multiplication is a binary operator and division is a binary partial operator on the real numbers. This is ...


2

What you're saying is right. The subobjects are not in [a reasonable, natural] bijection with the quotient objects; one way you could associate a quotient object $G \to Q$ to each subobject $H \to G$ is to take the quotient by the smallest normal subgroup generated by $H$, but this is not one-to-one.


2

One has to distinguish the property of being a monomorphism from the property of being the kernel of another morphism. These two are the same in any abelian category (by definition, and this allows for the usual correspondence between subobjects and quotient objects by taking kernels and cokernels), but in general, though any kernel morphism is a ...


2

The part about the opposite category is probably not so useful for thinking about these things, simply because that category is not very similar to the original category. You ask where the normal subgroups arise when speaking about subobjects. Well, they are precisely those subobjects of some group $G$, which are kernels (remember that a morphism $f: H\to ...


2

A good introductory reference explaining the relationship between the four models mentioned by Zhen Lin is the survey paper Julia Bergner, A survey of (infinity,1)-categories, in J. Baez and J. P. May, Towards Higher Categories, IMA Volumes in Mathematics and Its Applications, Springer, 2010, 69-83, pdf. A good idea would be first to study enriched ...


2

I guess you want to prove that, given a morphism $f: A\to B$ and a subobject $j: S\rightarrowtail B$ with cokernel $\pi: B\twoheadrightarrow\text{coker}(S\rightarrowtail B)$ (in an abelian category), you have $f$ factoring through $j$ if and only if $\pi\circ f=0$? For the direction "$\Leftarrow$" note that any monomorphism is the kernel of its cokernel. ...


2

The question is answered by Zhen Lin in the comment. Morphisms of subobjects (resp. quotients) are defined as morphisms of the underlying objects which are compatible with the inclusion morphism (resp. projection morphism). And of course Mac Lane mentions this.


2

$\mathrm{Mon}(C)$ does not have composition as a tensor. The composite of two monads is not a monad in any natural way unless there is a distributive law between them. Lots of research goes into getting around this fact, e.g. the paper you cite. For example, for any set $A$ there is a functor $\Delta_A: \mathsf{Set} \to \mathsf{Set}$ which is constant at ...


2

Well, any group object in $\mathbf{Ring}$ will be a group object in $\mathbf{Ab}$, and it is straightforward (using the Eckmann–Hilton argument) to show that every object in $\mathbf{Ab}$ admits the structure of a group object in a unique way: the new group operation will coincide with the old group operation. Thus any object in $\mathbf{Ring}$ has at most ...


2

In any category, an isomorphism of two objects $A,B$ is just a morphism $f$ from $A$ to $B$ so that there exists a two-sided inverse- a morphism $g:B\to A$ such that the composition in both directions is the identity. Applying the definition to the category of pointed sets gives us that an isomorphism in the category of pointed sets between $(A,a)$ and ...


2

I would say that the maps are naturally equivalent. You're right that saying naturally isomorphic is kinda abusing the terminology but it doesn't sound monstrous to me. Whatever you choose you should just make sure to explain it the first time you use it.


2

It is no abuse of notation to say that $F(f)$ and $G(f)$ are isomorphic: $F(f)$ is isomorphic to $G(f)$ in the arrow category of $D$.


2

The category $\mathcal{C}$ of all group actions is complete and cocomplete. First, observe that there is an evident forgetful functor $U : \mathcal{C} \to \mathbf{Set} \times \mathbf{Set}$ that preserves and creates limits. You can also check that $U : \mathcal{C} \to \mathbf{Set} \times \mathbf{Set}$ preserves and creates filtered colimits. With further ...


1

Suppose that $F$ takes s.e.s. of chain complexes of vector spaces to l.e.s. of vector spaces. Suppose that the objects of $F(0 \to A^{\bullet} \to B^{\bullet} \to C^{\bullet} \to 0)$ are $H^0(A)$, $H^0(B)$, $H^0(C)$, $H^1(A)$, etcetera and that the maps $H^i(A) \to H^i(B)$ and $H^i(B) \to H^i(C)$ are the standard ones. Suppose that $F$ commutes with direct ...


1

$F$ won't be covariant, but rather a contravariant functor. If $X \to Y$ is a morphism, this induces a natural transformation $- \times X \to - \times Y$, which in turn induces a natural transformation $\hom(-,R^Y) \cong \hom(- \times Y,R) \to \hom(- \times X,R) \cong \hom(-,R^X)$, i.e. a morphism $R^Y \to R^X$.


1

Let me try fixing context and notation first. So ${\mathscr C}$ is a Cartesian closed category, and if $f: B\times X\to Y$ is a morphism in ${\mathscr C}$, we denote $\lambda_X f: B\to Y^X$ the adjoint/$\lambda$-closure of $f$ with respect to $X$. Then, given $g: A\to B$, you're asking for a proof of $$(\lambda_X f)\circ g = \lambda_X (A\times ...


1

Any ring can be viewed as an additive category with one object and the endomorphisms of that object given by the elements of the ring. Then Gabriel-Zisman localization applied to this special case recovers the usual localization of a ring at a multiplicative subset of elements. This makes sense even when the ring is not commutative, and then one recovers ...


1

Petri nets model networks and distributed systems, but also chemical reactions. They can be seen as symmetric monoidal categories. The Azimuth Project, Petri net V. Sassone has several publications about this connection between Petri nets and category theory.


1

I can recommend the following introduction: Moritz Groth, A short course on infinity-categories, http://arxiv.org/abs/1007.2925


1

Well, for one, these functors are closed under composition; for example, if you have natural transformations $\tau : \mathbf 1 \to F, \eta : \mathbf 1 \to G$, you get natural transformations $\mathbf 1 \to FG$ and $\mathbf 1 \to GF$. You have natural transformations $\mathbf 1 \to F^n$ for all $n$. I like to think of a natural transformation $\mathbf 1 \to ...


1

Both the product and coproduct of a family of sets $X_i$ are given as the disjoint union $\coprod X_i$ of sets together with the relations $$\left\{(a, a) \in \left(\coprod X_i\right) \times X_j \ \middle| \ \forall a \in X_j\right\}$$ and $$\left\{(a, a) \in X_j \times \left(\coprod X_i\right) \ \middle| \ \forall a \in X_j\right\}.$$


1

By definition of the exponential object, $\hom(X^{A+B}, X^A) \cong \hom(X^{A+B} \times A, X)$. So you want a natural morphism $X^{A+B} \times A \to X$ (and similarly for $B$, then you take the product of these two morphisms). But again by definition of the exponential object you have a natural morphism: $$\operatorname{eval} : X^{A+B} \times (A+B) \to X.$$ ...



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