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0

The first is not misleading, the notation $(v, w)$ does not imply product vs. coproduct, it's just notation for a tuple. If we had infinitely many summands (so that product $\neq$ coproduct) then writing $(v_i)$ just indicates that you have a tuple indexed by $i$. It does not tell you whether that tuple has finite support or not and when it does have ...


2

The quasi-isomorphisms in the category of chain complexes are related to homotopy theory, and in a sense localizing at the quasi-isomorphisms is like localizing topological spaces at the homotopy equivalences. It is primarily a computational tool for chain complexes and resolutions of various objects. The purpose achieved by localizing, in general, is that ...


1

Well, as other sources and the comments suggests, by proper usage of the universal properties, one can avoid speaking about natural transformations such as $\def\im{{\rm im\,}}\def\tto{\dashrightarrow} \im f\tto \ker g$ for exact sequences $\overset f\to \overset g\to$. However, probably the following lemma and preliminary setting can be useful to exactly ...


1

Actually, you have an error in understanding the category $\mathcal{G}$ obtained from $G$. It is not the full subcategory of $\mathbf{Grp}$ whose only object is $G$: i.e. it is not the category with one object whose arrows are endomorphisms of $G$. Instead, it is the abstract category with one object whose arrows are elements of $G$. The product of arrows ...


0

Very quick answer, since i am on a satelite connection (very expensive). Here you have an interactive tool for the category of finite sets. Wildcats is my (free) package for Mathematica which can be used to do some calculations and visualizations in category theory.


1

The fonction $x \mapsto 1/x$ defined on $[1,+\infty[$ is uniformly continuous. (Exercise !) The function $x\mapsto 1/x$ defined on $]0,1]$ is continuous but not uniformly continuous (exercise again !) and these two functions are inverse one to each other. As on a metric space, with the uniform structure given by the metric, being uniform amouts to be ...


0

The wonderful "and so on" paper by Kelly answers almost any possible question on the subject of pointed functors: Max Kelly, “A unified treatment of transfinite constructions for free algebras, free monoids, colimits, associated sheaves, and so on.” Bull. Austral. Math. Soc. 22 (1980), 1–83. A dry account of what Kelly does: in the lab, where else?


1

Let $V_i$ and $G_i$ be the vertex and edge set of $G_i$ respectively. Note that $V_i\subset V(G)$ and $E_i\subset E(G)$ where $V(G)$ and $E(G)$ are the vertex and edge sets of $G$ respectively. Let $v_i\colon V_{i+1}\to V_i$ and $e_i\colon E_i\to E_i$ be the maps induced by $f_i$ on these finite sets in the obvious way. It's a general property of finite sets ...


2

No, for example if your diagram is just a single arrow $A \to B$ then $J$ will be a category with two objects, say $\{1, 2\}$ and one non-identity homomorphism $1 \to 2$. Then the diagram $A \to B$ is given by the functor $J \to C$ which sends $1 \mapsto A$, $2 \mapsto B$, and $(1 \to 2) \mapsto (A \to B)$. Here $C$ could be anything, sets, groups, top, ...


7

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 ...


3

$\mathrm{Mon}(C)$ 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

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 ...


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 $F \to \hom(-, L)$. ...


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 $$\left\{(a, a) \in X_j \times \left(\coprod X_i\right) \ \middle| \ \forall a \in X_j\right\}.$$


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.


4

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 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 ...


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 ...


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

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

$F$ 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 $R^Y \to R^X$.


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. ...


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 ...


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 ...


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 ...


1

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


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

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

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

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

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.


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 ...


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.


0

In the programming language Haskell, the category Hask is the category with Haskell types as objects and functions between types as values. A functor from Hask to Set is one that takes each type to a set of each possible value to that type and functions to functions. This is concrete.


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

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 ...


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 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 $X$. ...


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: $$\operatorname{eval} : X^{A+B} \times (A+B) \to X.$$ ...


0

The words "dispensible" and "merely a representative" are just informal descriptions of the definition of a natural transformation. Therefore you won't be able to find a "proof" from the definition. The definition says what it says and it cannot really be simplified. Another informal description is that the $\eta_A$ vary "continuously with $A$". This is ...


0

A covariant exact functor ${\bf A}\to{\bf B}$ preserves finite limits and colimits, so a contravariant exact functor does it too as a (covariant) functor on the opposite category: $F:{\bf A}^{op}\to {\bf B}$. Here ${\bf A}^{op}$ is again Abelian (as [at least one form of] the axioms are self dual). Now limits (in particular, kernels and products) in ${\bf ...


1

It contains identities. It is closed under composition (the composition of two surjective functions is also surjective). You can find two objects $A,B$ in $\bf Set$ and a morphism $A\to B$ which doesn't belong to this subcategory of surjectives, but both $A$ and $B$ do.


0

You should just use the universal property, don't worry about exact sequences. For starters assume the $p_i$ exist. You have to show that a family of maps $N_i\colon X_i \to Z$ factors through a unique map $X \to Z$. Well using the $p_i$ you get maps $X \overset{p_i}{\rightarrow} X_i \overset{N_i}{\rightarrow} Z$ so sum this over $i$ to get $X \to Z$. ...


1

As discussed, it suffices to prove the claim for $\mathcal{E} = \mathbf{Set}$. Let $x_1 \in X_1$, $y \in Y$, and suppose $\pi (x_1) = f (y)$. Consider $y$: Either $y = \pi (y_1)$ for some $y_1 \in Y_1$, or $y = \pi (y_2)$ for some $y_2 \in Y_2$, in which case $\pi (x_1) = f (\pi (y_2)) = \pi (f_2 (y_2))$; but that implies $f_2 (y_2) = k_2 (x_0)$ for some ...


4

The universal property of the product $X \times Y$ is stated in terms of two projection maps $p_X : X \times Y \to X$ and $p_Y : X \times Y \to Y$. Any map $X \to Z$ can be composed with the projection $p_X$ to get a map $X \times Y \to Z$, which does exactly what you'd expect: it only looks at the $X$ variable and it ignores the $Y$ variable. In addition, ...


2

Let $\mathbf{Qcoh}_R$ be the total category of modules: The objects are pairs $(A, M)$ where $A$ is a commutative $R$-algebra and $M$ is an $A$-module. The morphisms $(A, M) \to (B, N)$ are pairs $(f, g)$ where $f : A \to B$ is an $R$-algebra homomorphism and $g : M \to N$ is an abelian group homomorphism such that $g (a m) = f (a) g (m)$. Composition and ...


5

One quick answer is that the study of abelian groups is the same as the study of $\mathbb Z$-modules, while the study of groups is not at all the study of modules over a principal ideal domain (or any other ring). The theory of modules over a principal ideal domain is very down-to-earth, with a very nice representation theory, rank theory, etc.. Thus, the ...


2

The question is quite interesting, at least from an epistemological point of view. It is not easy to work out an all-comprehensive answer which is in the meantime meaningful enough in a broad context, but I will try to say something that hopefully can be understood, at least on a metamathematical level, whatever this may mean. Also, since you have tagged the ...


0

Your idea is not quite right: the product $M\times N$ of two modules $M,N$ is the pullback of $M\to 0\leftarrow N$, but in your problem you are interested in the pullback of ${\mathbb Z}_{p^2}\to {\mathbb Z}_p\leftarrow {\mathbb Z}$, which can be realized as the submodule of ${\mathbb Z}_{p^2}\times{\mathbb Z}$ consisting of those pairs $(\overline{x},y)$ ...



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