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## New answers tagged category-theory

0

Consider a bigger category $\mathcal D$ which disjointly contains $C^\downarrow$ and $C^{\downarrow\downarrow}$, and additionally it contains the arrow $g$ of $C$ as an arrow from $f$ in $C^\downarrow$ to $(u,v)$ in $C^{\downarrow\downarrow}$ whenever $fgu=fgv$. All compositions are defined straightforwardly. Now, verify that in $\mathcal D$, the ...

0

Written equalities are just confusing when dealing with categorical properties. Prefer diagrams to it. Suppose $F,F' \colon A \to B$ admits both $G \colon B \to A$ as right adjoint. By definition, there are bijections $$\iota(x,y) \colon \hom_A(x,Gy) \simeq \hom_B(Fx,y) \\ \iota'(x,y) \colon \hom_A(x,Gy) \simeq \hom_B(F'x,y)$$ natural in $x$ and $y$ (i.e. ...

1

The 2-out-of-3 property is a bit too weak for this. You can deduce it from the 2-out-of-6 property, but unfortunately that isn't given. Instead, you have to use the notion of asphericity (and axioms 4 and 5): see corollary 7.3.11 and proposition 7.3.12 in my notes.

1

Another way to put it. An abelian category $\mathscr M$ is in particular a category enriched over the category $\mathsf{Ab}$ of abelian groups. So for two objects $A,B$ of $\mathscr M$, the hom-set $\hom_{\mathscr M}(A,B)$ actually carries a structure of abelian group : in particular, it has a neutral, which is the zero map from $A$ to $B$.

0

Any object $C \in \mathcal{C}$ can be seen as a constant functor sending everything to the terminal category containing $C$, so yes, the objects of $\mathcal{C}$ are being sent to the objects of $\mathcal{D}$. And any morphism in $\mathcal{C}$ is a natural transformation of constant functors (easy exercise). So your $a$ gives rise to a functor $a: ... 1 The answer of Amr focuses on connected groupoids. In a certain sense, a connected groupoid is not significantly different from it's automorphism group. However, the very reason why groupoids are more fundamental than groups is that they can have more than one connected component. One interesting construction enabled by having more than one component is to ... 0 Is not the best way of defining the structure sheaf. It just uses the fact that$D(f)$form base for the topology. To see the last isomorphism if you want you can define the sheaf of regular functions$\mathscr{R}(U)=\cap_{x\in X}A_{\frak{p}_x}$. The correct way of understanding this is that for each open it gives you the ring of regular functions on$U$and ... 0 Let me summarize the information from the comments and links into a concrete answer. Yes, a homomorphism is expected to be a structure-preserving map. This also applies to homomorphisms of relational structures. It is possible to think of the objects in a concrete category as sets, and of the morphisms as certain maps between those sets. There are ... 3 The answer$A \cong S^2 \vee S^1$isn't correct. Notice that if you remove the collapsed point from$A$, then you get a space homeomorphic to$(0,1) \times S^1$. There is no point of$S^2 \vee S^1$that has this property (no matter what, you'll still be left with a part of dimension one). Your space is actually homeomorphic to a sphere with two points ... 1 Since the "morphism category" (the more usual name is "arrow category" though) of$C$is nothing but a special type of comma category - the category$\left(\text{Id}_C\downarrow \text{Id}_C\right)$- this question has been exaustively answered a long time ago by myself and others here and more recently here. In short: what you read in most/all category ... 0 I somehow find the solution myself. The example I gave is just wrong, it is not middle linear. (Thanks to @Jeremy Rickard for noticing.) And since the identity of morphism is unique (Thanks to @roman for a remind),$1_{f}$must be the identity isomorphism on$C$. And similarly for$1_{g}$. Hence for$h$to be an equivalence, there must exist$h'$, such ... 3 From the way I interpreted your question, my answer would be that I don't think there is anything interesting you can do with a connected groupoid other than looking at the size of the set of objects and the automorphism group of one of the objects. A connected groupoid is determined up to isomorphism by these two properties. Any other property you will ... 2 No, there is no category of all categories. Likewise, there is no "set of all sets" or "class of all classes" or..."mother of all mothers". Think about it: the mother of all mothers, if it existed, would be a mother (by definition), so she would have to be the mother of herself. Pretty akward. Nevertheless, in mathematics it is important to be able to ... 2 The category cannot be abelian, because a mono-epi is not necessarily an isomorphism. Take any (Hausdorff abelian) topological group$G$and consider$G_d$, the same group but with the discrete topology; the identity map$G_d\to G$is certainly mono and epi, but it's not an isomorphism unless$G$was discrete to begin with. For the kernel, the usual one ... 2 The zero morphism$A \to B$can be factored into $$A \to 0 \to B$$ where$0$is a zero object. (i.e. it is a terminal object and an initial object) As an aside, when you wrote it as "$0$-morphism", my first reaction was that you were referring to the concept from higher category theory; e.g. in$\mathbf{Cat}$, categories are$0$-morphisms (i.e. ... 0 The universal property is defined via the forgetful functor:$\mathrm{U}:\mathrm{Alg}\to\mathrm{Vec}$This defines its adjoint functor:$\mathrm{T}:\mathrm{Vec}\to\mathrm{Alg}$3 No, the extended sequence needn't be exact. Take first$0\to A\to B\to D\to 0$exact so that$0\to \hom(M,A)\to \hom(M,B)\to\hom(M,C)\to\operatorname{Ext}^1(M,A)$is exact. Take any$E$such that$\hom(M,E)\neq 0$. Then set$C=D\oplus E$.$0\to A\to B\to C$is exact, but then the sequence$0\to \hom(M,A)\to\hom(M,B)\to\hom(M,C)\to \operatorname{Ext}(M,A)$... 0 It seems to be a simple consequence of the fact that$U\colon \bf Top\to Set$admits a left adjoint$F$which sends a set$X$in the discrete topological space$FX$(it must be what you mean by "initial topology"). So a simple way to characterize the lift of$f_i\colon X\to UA_i$is via the bijection $${\bf Set}(X, UA)\cong {\bf Top}(FX,A)$$ sending ... 1 I don't think$Y^X$is semilocally simply connected when$X$isn't compact. The argument below is incomplete in that it uses the standard compact open topology on$Y^X$, while one would at least have to$k$-ify it, and probably do some more to stay inside the category you are considering. Nevertheless, I think it shows that semilocal simple connectedness ... 0 From my very limited knowledge of mathematical biology I was under the impression that the more relevant areas of mathematics will be the "applied" ones, as such I don't think that reviewing the abstract areas of mathematics will be very helpful too you. My other thought is that without doing a maths degree it will be very difficult to cover much ground in ... 4 The Yoneda lemma is a very powerful tool, use it. The definition of an exponential object in a category$C$is (where this is a natural isomorphism in$U): $$\hom_C(U \times V, W) \cong \hom_C(U, W^V)$$ So here (I suggest you carefully check for each step which property I used): \begin{align} \hom_C(U \times V, W) & \cong \hom_C(U \times V, GF(W)) \\ ... 5 You can consider pretty much the same thing; e.g. given a set X of arrows, you can consider the set of arrows \{ x \circ f \mid x \in X \} , or the reverse, or a two-sided thing. A particularly notable special case is that of a sieve, where we consider the set generated as above, but all of the arrows of X have codomain to be a single object of the ... 4 I know at least two, maybe even three contexts where "derivatives" (in a loose sense) exist: If F : A \to B is a (covariant) functor between abelian categories which is left exact, and the categories in question are well behaved, then we can talk about the right derived functors R^iF of F. These are functors such that whenever 0 \to A \to B \to C ... 0 A categorical presentation of a notion of derivative for polynomial functors (aka containers) generalizing Huet's zipper is given in ∂ for Data: Differentiating Data Structures. An explanation of this work in term of programming constructs is given by McBride in Clowns to the left of me, jokers to the right. 1 The idea it's correct but personally I find more easily to follow the argument when stated in the following way. More in details it should work like this: by Yoneda lemma we have that every natural transformation of kind\tau' \colon \hom_{\mathcal D}(-,r) \Rightarrow \hom_{\mathcal D}(-,r')$$is of the form \hom_{\mathcal D}(-,h) for some h \in ... 4 If F : \mathcal{C} \to \mathcal{D} is essentially surjective on objects, then the induced functor F^* : [\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}] is conservative, but not necessarily fully faithful. F^* : [\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}] is an equivalence of ... 2 Automorphisms of the lattice L={\mathfrak T}(X) of topologies on arbitrary set X are classified as follows (see here): If X is infinite or has cardinality \le 2, then every automorphism \phi of L is induced by a bijection f_\phi: X\to X. In particular, we obtain a natural homeomorphism of the topological spaces$$ f: (X,\tau)\to (X, ... 0 A categorical productX\times Y$must satisfy the universal property that for each object$Z$and morphisms$f$and$g$, there exists a unique morphism$h$such that the following diagram commutes $$\begin{matrix} & &Z\\ &\swarrow^f&\downarrow^h&\searrow^g\\ X&\leftarrow^p&X\times Y&\rightarrow^q&Y \end{matrix}$$ If we ... 9 (Tyler answered in a comment after I started writing this. I'm posting it because of the extra information I give) Yes-ish. I mean, you can do it, but for arbitrary topological spaces there is a difference between homotopy equivalence and weak homotopy equivalence, and you need to choose which one you care about. The construction is well known, and goes as ... 0 There are many way to prove this, but I think the easiest (in some sense) is via yoneda embedding and using the definition of limits for a functor$D \colon \mathbf I \to \mathbf C$as terminal objects in the category of cones over$D$. We have that for every category$\mathbf C$there's an embedding$y \colon \mathbf C \to [\mathbf ...

2

A covariant functor $F$ from $\tilde S$ ($S$ viewed as a category) to $\tilde T$ obviously sends the unique object of $\tilde S$ to the unique object of $\tilde T$. It maps "morphisms" in $\tilde S$ (that is, elements of the monoid $S$) to "morphisms" in $\tilde T$, and therefore gives a function $f : S \to T$. Finally, the functor satisfies $F(a \circ b) = ... 2 From$a \circ h = id_A$you can deduce from the functoriality of$F$that$F(a) \circ F(h) = id_{F(a)}$. But by definition of$h$you have a commutative square: $$\begin{array}{ccc} F(A) & \xrightarrow{a} & A \\ \downarrow F(h) & & \downarrow h \\ F(F(A)) & \xrightarrow{F(a)} & F(A) \end{array}$$ Therefore$h \circ a = F(a) \circ F(h) ...

5

The most important (and in fact, motivating) example is the category of sheaves on a space $X$. Here a global element is the same as a global section (this also explains the name). Of course, not every sheaf is defined via its global sections, but they are definitely interesting. In fact, one of the most important theorems in a first course on algebraic ...

2

The question only makes sense when you specify the category in which you are working. For an algebraic geometer the most natural choice would be the category of varieties (or even better, schemes) over some field $k$. Then the product $\mathbb{P}(V) \times \mathbb{P}(W)$ embeds into $\mathbb{P}(V \otimes W)$, via the Segre embedding mapping a pair of lines ...

2

In the case $\mu$ is a natural isomorphism, the monad is said to be idempotent. In this case, the forgetful functor from the Eilenberg-Moore category of algebras is full and faithful.

0

The question has been answered on mathoverflow.net. The equality I forgot to use was that $c_{I,A} = 1_A$.

2

The usual notion of equivalence for model categories is due to Quillen: given model categories $\mathcal{M}$ and $\mathcal{N}$, an adjunction $$F \dashv G : \mathcal{N} \to \mathcal{M}$$ is a Quillen equivalence if it is a Quillen adjunction (i.e. $F$ preserves cofibrations and trivial cofibrations, and $G$ preserves fibrations and trivial fibrations) such ...

1

It may be useful to see an example on a more interesting category. Let $X$ be a topological space, and consider the category of all bundles over $X$. Recall that a bundle over $X$ is the same thing as a continuous map $E \to X$; $E$ is called the "total space" of the bundle. A morphism of between two bundles $E \to X$ and $E' \to X$ is a continuous map $E ... 1 "Surjective" and "epimorphic" have the same meaning. They specifically mean that every element of the range occurs as the image under$f$of some domain element. Neither of those has anything to do with "total". The term "total function" means exactly the same thing as "function". The term "total function" is used to imply a contrast with "partial ... 1 Suppose that$\{D_i, \psi_i^j\}$is your system, and for$i \leq j$you have a map$\psi_i^j : D_j \to D_i$, and let$\alpha_i : lim D_i \to D_i$be the maps of the limit (the ones with the universal property). Then you have for every$X$an isomorphism: $$\begin{array}{cccc} \theta_X : & Hom(X, d) & \to & lim Hom(X,D_i) \\ ... 3 \mathbf{Set}^\mathrm{op} is not cartesian closed, because 1 + X \not\cong 1 in general. (In a cartesian closed category, 0 \times X \cong 0.) 6 What do you mean by "e(1)"? The identity in your group is the morphism e itself. In any category C with a terminal object 1, a morphism 1 \to c for an object c \in C is called a global point of c. In some, but not all, categories, this often gives a reasonable "underlying set" functor \text{Hom}(1, -) : C \to \text{Set}, although it may not ... 1 1.- You can not talk about the "identity" of G, because G may not be a set. But thats the intuition. 2.- 1 is a final object because thats the definition, not a consequence. The morphism G \to 1 is unique because 1 is a final object. 3.- The two assertions are equivalent. A category has finite products if and only if has the product of every two ... 11 I think one can give not just one but several intuitive descriptions of adjunctions, some of which are more appropriate for understanding some adjunctions than others. For this reason it is perhaps less useful to ask someone else to describe their intuitions than to just collect a list of examples and build your own intuition from that list. But let me try ... 4 In general, I suspect an adjunction whenever a functor preserves limits of colimits. I then try to find the adjoint or to apply Freyd's theorem. Of course, sometimes I fail but as "adjoints are everywhere" I also often win. But there is a particular case when I find it easy to see adjunctions. Take a category \mathsf C and a full subcategory \mathsf D ... 2 My personal favorite description uses profunctors. If \mathcal A and \mathcal B are categories, a profunctor F:\mathcal A\not\to\mathcal B "from \mathcal A to \mathcal B" is defined as a functor F:\mathcal A^{op}\times\mathcal B\to\mathcal{Set}. The bigger category that disjointly contains \mathcal A and \mathcal B and the sets F(a,b) are ... 7$$\hom(R/I \otimes S,-) = \hom(R/I,-) \times \hom(S,-) = \{f \in \hom(R,-):f(I)=0\} \times \hom(S,-) = \{h \in \hom(R \otimes S,-) : h(I \otimes 1)=0\}=\hom((R \otimes S)/I^e,-).$$Hence,$R/I \otimes S = (R \otimes S)/I^e$. Then also$R/I \otimes S/J = (R \otimes S/J)/I^e = (R \otimes S)/(J^e,I^e)$. 2 Every single nonempty set can be made into a pointed set by choosing out an arbitrary element. A collection of infinite number of nonempty set, however, need not be capable of all being made into pointed set, if Axiom of Choice is not accepted. Consider an index set$I$that is infinite. Let$A_{i}$be a collection of nonempty set, and let$B_{i}$be a ... 5 There is a functor from the category of pointed sets to the category of sets, that is, forgetting the chosen point. However there's no “canonical” functor backwards unless you define it by choosing anything which is not a set and add it. Actually the category of pointed sets is equivalent to the category of sets with partial maps, which is not equivalent to ... 4 A non-empty set is not a pointed set until you actually pick a point in it. The pointed set is the combination of a set and a particular element in it. So$\{1,2\}$is not a pointed set, but$\langle \{1,2\},1\rangle$and$\langle\{1,2\},2\rangle\$ are two different pointed sets that can be made from it. And if somebody talks about "pointed sets" rather ...

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