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I have at least managed to construct a left adjoint to the associated bundle functor $$P [-]: \mathsf{Space}_G \to \mathsf{Bund}(M),$$ Given by the "fibre product" functor $$P \times_M (-): \mathsf{Bund}(M) \to \mathsf{Space}_G,$$ where the $G$ action is the obvious one $(u_x, v_x).g = (u_x.g, v_x)$. The counit is given by the $G$-equivariant maps $$\... 1 Thanks to Zhen Lin for his comment, here is what I believe is the correct interpretation of the words and phrases having been unclear to me. i) The pullback of the morphism f_A:A \to C along the morphism f_B:B \to C is the morphism p_B:P \to B where \langle P, p_A, p_B \rangle is the pullback, i.e. f_A \circ p_A = f_B \circ p_B where P is ... 0 If there is "co-exponential" f:B\to C\times A as you suggest, let f_1:B\to C and f_2:B\to A be the compositions of f with the projections. The universal property of f says that given any g=(g_1,g_2):B\to D\times A, there is a unique h:C\to D such that g=(h\times 1)f. Composing with the projections, g=(h\times 1)f just says that g_1=hf_1 ... 0 As discussed in the comments, this almost always fails. In particular, it fails in your example (as long as V is nonzero): if \alpha:k\to k is any automorphism, then you can define a ring-homomorphism \varphi:k[x,y,z]\to \textrm{End}_k(V) by \varphi(f(x,y,z))=\alpha(f(0,0,0)) (where \alpha(f(0,0,0))\in k acts on V by scalar multiplication). ... 1 In universal algebra this concept is also known as the kernel of a morphism f : X\to Y. However, topological spaces are not universal algebras and calling these things "kernels" is not a good idea, because the term is reserved for a better known concept already. Incidentally, in a typical "concrete category of sets with some structure", like \mathsf{Top}... 0 The problem in question (a) is stated (by your professor I guess) in a terrible way... The given data of the problem is the G-sets X and Y and the points x \in X and y\in Y : it is then asked to find a property \mathsf P depending on those data such that \mathsf P is equivalent to the property "\exists f \colon X \to Y morphism of G-sets such ... 1 Looking at x \in X as a map x \colon 1 \to X, the operator \tilde x is precisely the natural transformation given by co-Yoneda's embedding$$ \mathsf{Set}(X,-) \stackrel{\mathsf{Set}(x,-)} \longrightarrow \mathsf{Set}(1,-) \simeq \mathrm{id}_{\mathsf{Set}} $$So you could call it the restriction along x, or precomposition by x, and denote it x^\... 1 By definition, a functor G : C\to D is full and faithful iff C(c,c')\cong D(Gc,Gc'). Again by definition, a functor is codense iff \text{Id}_D\cong \text{Ran}_FF, i.e. iff d\cong \int_c Fc^{D(d,Fc)} (coends, p. 16), naturally in d\in D. Let's put these two definitions together to prove the result: the only additional assumption I need is that D ... 3 Such a morphism f : a \to b is said to be F-(hyper)cocartesian. This is in connection to Grothendieck opfibrations. You may like to work out what this means concretely in the case of \mathrm{dom} : [\mathbf{2}, \mathcal{C}] \to \mathcal{C} where \mathcal{C} is a category with pushouts. 2 This was alluded to in the comments and may not be what you're looking for, but it surely deserves mention that you can take \mathbf{T} to be the category of compact Hausdorff spaces. The category \mathbf{Ab}(\mathbf{T}) is the the category of compact abelian groups, which is equivalent to \mathbf{Ab}^{op} and hence abelian by Pontryagin duality. 2 A split epi in \mathbf{Top} need not even be a fiber bundle at all: its fibers don't have to all be homeomorphic. For instance, consider the map p:[0,1]\to[0,1] given by p(x)=\min(2x,1). This is a split epi (it is split by x\mapsto x/2), but p^{-1}(\{1\})=[1/2,1] is an entire interval while the other fibers of p are all single points. 1 Stone spaces are not cartesian closed, and I don't know of any interesting subcategory of compact Hausdorff spaces that is. Typically, you get an interesting cartesian closed category related to compact Hausdorff spaces by allowing more spaces, not less: the problem is not that your spaces are too general, but simply that the natural topology on the set of ... 2 Suppose you have a map f:Y\to Z such that (\operatorname{coker}\phi,\operatorname{coker}\psi)\circ f=0. Then \operatorname{coker}\phi\circ f=0 and \operatorname{coker}\psi\circ f=0, so f factors through \ker(\operatorname{coker}\phi)=\phi and \ker(\operatorname{coker}\psi)=\psi. It follows that f factors uniquely through the pullback as ... 2 In linear algebra the double-dual comes to mind, but the general term that's most relevant is "evaluation map". One somewhat common notation is \mathrm{ev}_{x}(f)=f(x). Either ev (your ~) is the evaluation map, or each \mathrm{ev}_{x} is the evaluation map at x. 2 In the context of functional analysis, the map$$ \Phi: A \to (\Bbb F)^{\Bbb F^A}\\ \Phi:x \mapsto (f \mapsto f(x)) $$is called the "evaluation map". 5 It's called the canonical map into the double dual. It does not have another name that I know of. In a monoidal category with duals (I'm being vague), the uncurried version of it, \operatorname{ev} : X \otimes X^* \to I, is called evaluation. 0 Here is how I do it. Sometimes limits are called inverse limits and denoted by \varprojlim, whereas colimits are called direct limits and denoted by \varinjlim. Now, a functor which has an adjoint in some direction, preserve the limits in the same direction. Unravelling, a functor which has a left adjoint (i.e. which is a right adjoint) preserves ... 3 The constant sequences (c_n)_{n\in\Bbb{N}} of \Bbb{N} are not in the inverse limit because they do not satisfy$$\theta_{n\leftarrow n+1}(c_{n+1})=c_n,$$for any n at all. To see that the inverse limit is empty, suppose toward a contradiction that (x_n)_{n\in\Bbb{N}} is in the inverse limit. Then it satisfies x_n=\theta_{n\leftarrow n+1}(x_{n+1}) ... 5 The problem is that as the indices increase the numbers decrease by 1. Eventually we will have to hit a negative number, but \mathbb N does not contain any of those. 5 Yes, that's right. This perspective leads to the idea that linear categories are a "many-object" generalization of rings, and naturally occurs in Morita theory. 3 Let \mathcal{C} be a category and U : \mathcal{C} \to \mathsf{Set} be a functor with left-adjoint F. If there is an A\in \mathcal{C} with |UA|\geq 2, then the unit of the adjunction F \dashv U is componentwise injective (see this answer). Now observe, that every monad T on a category \mathcal{A} arises from some adjunction between \mathcal{... 4 Let f : A \to B be a function. We have a commutative diagram$$ \begin{matrix} A &\xrightarrow{\eta_A}& T(A) \\ {\ \ } \downarrow {f} & & {\quad}\downarrow {T(f)} & & \\ B &\xrightarrow{\eta_B}& T(B) \end{matrix} $$In particular, if x,y \in A satisfy \eta_A(x) = \eta_A(y), then \eta_B(f(x)) = \eta_B(f(y)) However, ... 5 This question is not really about category theory itself (though category theory is the first subject in which the issue you are running into cannot be easily swept under the rug). 1. and 2. could be equally well asked of set theory and basic algebra "In what way does the collection of all sets consist of sets?" "What are collections of sets actually?" ... -1 Some thoughts from a definitely non expert: When doing mathematics, you have to first choose a logical system within which to work. The first axiomatic system introduced to mathematicians is practically always ZFC, but by no means is it the only choice. Category theory is interested in the consequences of composition of events, and so it only defines ... 1 To say that a category "consists of the data" means that in order to specify a category, one has to tell their reader exactly the data that lies inside the definition. For instance, a monoid consists of the following data: A set M; a binary operation \circ:M \times M \to M that is associative, i.e., \circ(g,\circ(h,k)) = \circ(\circ(g,h),k); and a ... 0 A category consists of something in much the same way a group consists of an operation and a set. Often a category is a tuple of a class of objects and a class of arrows. However because sometimes categories deal with collections that are larger then classes it makes sense to use a vague term that basically means "Use the appropriate set/class/etc concept ... 2 Morphisms of G-actions are the same as morphisms of G-sets. To clarify this, I need to introduce the necessary definitions. By right G-action (on set X) we usually mean a function \cdot\,\colon X\times G\to X such that x\cdot(gh) = (x\cdot g)\cdot h and x\cdot e_G = x. Morphism between G-actions on sets X and Y are defined as functions ... 2 This is just a standard abuse of notation that you are surely familiar with in other contexts. A group is not just a set G, but actually a pair (G,\cdot) where \cdot is a binary operation on G, but we nevertheless frequently refer to "G" alone as the group. Similarly, a right G-set is really a set X together with an action X\times G\to X, ... 0 Setup Following page 32 of Vakil's notes,let S be a multiplicative subring of a ring A; i.e., 1 ∈ S ∧ x,y ∈ S ⇒ x · y ∈ S. Then we consider “formal fractions”, S⁻¹A ≔ { a / s ∣ a ∈ A , s ∈ S } The property we're interested in is 𝒫 : A-algebra → Bool 𝒫 f ≔ for every e in S, f e ∈ B is invertible Want to show: S⁻¹A is initial among A-algebras B ... 1 This analogous to the adjuntion between the singular simplicial set functor and geometric realization, or to the adjunction between the nerve functor and the fundamental category functor. Let I : \Delta \to \mathrm{Cat} be the "inclusion" that sends [n] to the category also called [n] that you mentioned in your description of \mathrm{str}. Then \... 1 In answer to a question that came up in comments under Najib's answer, let me point out that the category of pseudotopological spaces is a locally small cartesian closed category that contains \text{Top} as a full subcategory. Actually, this category \text{PsTop} has the even stronger property of being complete and locally cartesian closed and indeed is ... 1 If the product of B and C exists, then there is a natural bijection$$ \hom(A, B) \times \hom(A, C) \cong \hom(A, B \times C) $$We can write the function in the forward direction as (,): that is, it is the function that takes two morphisms g : A \to B and h : A \to C and produces the corresponding morphism (g,h) : A \to B \times C. On X \... 1 If a category \mathcal{A} has products, then there is a functor:$$\times : \mathcal{A} \times \mathcal{A} \to \mathcal{A}$$which maps a pair of objects (A,A') to A\times A' and where (f: A \to B, f' : A' \to B') is mapped to a morphism f\times f' : A\times A' \to B\times B', which you get by the universal property of B\times B' applied to the ... 4 You want to use the universal property of products, so you need arrows from X \times Y to X and Z, which will then give you an arrow from X \times Y to X \times Z. The first is the projection onto X and the second is the projection onto Y followed by the given arrow. 2 Let \mathbf{2} be the category with two objects and one morphism between them. Then the arrow category of any category \mathbf{C} is isomorphic to the functor category \text{Funct}(\mathbf{2}, \mathbf{C}); consequently, limits and colimits can be computed from the general facts about limits of functors. In particular, if \mathbf{C} has all (co)... 0 One has$$ p_2( i_1p_1+i_2p_2)=p_2i_1p_1+p_2i_2p_2=0p_1+1p_2=p_2=p_21_A $$and$$ p_1( i_1p_1+i_2p_2)=p_1i_1p_1+p_1i_2p_2=1p_1+0p_2=p_1=p_11_A. $$Hence, by uniqueness of the morphism f\colon A\rightarrow A such that p_1f=p_1 and p_2f=p_2 one has$$ i_1p_1+i_2p_2=1_A. $$4 What you actually have are two functors C^{op} \times D \to \mathsf{Set}, namely$${\rm Hom}_C(-, \mathcal{G}(-))$$and$${\rm Hom}_D(\mathcal{F}(-), -)$$And you want a natural isomorphism between them. 1 Yep, that's exactly what it means. 6 Note that taking opposite category gives rise to a 2-endofunctor:$$(-)^{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}...

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Given a morphism of short exact sequences, you get a morphism of long exact sequences, and this map respects composition. $\delta$ itself is, most precisely, a natural transformation between the functors $H_n\circ t$ and $H_{n-1}\circ f$ from short exact sequences of chain complexes to abelian groups, where $t$ sends an s.e.s. to its third complex and $f$, ...

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Yes, this is true. The point is that equivalence relations are the same thing as "equality after applying some function". In particular, there is an equivalence relation $\sim_f$ defined by $x\sim_f y$ if $f(x)=f(y)$. This equivalence relation contains all your pairs of points, so by definition it contains the minimal equivalence relation (call it $\sim$) ...

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Let $\mathcal{A}$ be a category of algebras over a monad over a regular (Barr-exact) category $\mathcal{X}$ in which regular epis split. Then $\mathcal{A}$ is also regular (Barr-Exact). In particular, assuming the axiom of choice, $\mathcal{A}$ is Barr-exact, if $\mathcal{X} = \mathsf{Set}$. A morphism is a regular epi, if it is the coequalizer of a pair ...

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You can use an injective resolution for $N$: let $E$ be injective and $0\to N\to E\to E/N\to 0$ be exact. Then the long exact sequence $$\DeclareMathOperator{\E}{Ext}\DeclareMathOperator{\H}{Hom} 0\to \H_R(\bigoplus_{i\in I}M_i,N)\to \H_R(\bigoplus_{i\in I}M_i,E)\to \H_R(\bigoplus_{i\in I}M_i,E/N)\to\\ \E_R^1(\bigoplus_{i\in I}M_i,N)\to \E_R^1(\bigoplus_{i\... 1 The proof is correct. If (C_i)_{i\in I} is any family of cochain complexes, then by writing out the definitions you immediately see H^n(\prod C_i)=\prod H^n(C_i). For if D denotes the differential on \prod C_i, d_i the differential on C_i, then H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i). 3 The sequence 0\rightarrow Mor(X,E')\rightarrow Mor(X,E)\rightarrow Mor(X,F) consists of of abelian groups, not of objects in the original category. So Lang is referring to the ordinary definition of exactness of sequences of maps of abelian groups. 4 It looks like you're trying to give a definition of the category of spans in the category of (finite) sets. In this case, composition will be given by pullback. Explicitly, given spans$$\begin{matrix} && A && \\ & {}^{f}{\swarrow} && {\searrow}^g \\ B &&&& C \end{matrix} \quad \text{and} \quad \begin{matrix} &...

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There are (at least) two different sort of things you can do, roughly corresponding to syntax and semantics. You can study the internal logic of a category, which lets you use a logical language as a convenient tool for doing calculations with in a category. e.g. objects are types, propositions are subobjects, et cetera. Conversely, given a suitable ...

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The underlying idea is extremely important in analysis; a simple example is that you can embed any inner product space into the space of linear functionals on its conjugate by the linear transformation $v \mapsto \langle -, v\rangle$. It's not an application of Yoneda lemma, but it's clearly the same sort of idea. In fact, there's a good analogy between ...

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yes $f$ factorized via the quotient for it the relationship you have defined on the points is finer than that associated to $f$ ie $xRy\Rightarrow xR_fy$ where $R_f$ the equivalence relation associated to$f$, $R_f$ is defined as: $xR_fy$ iff $f(x)=f(y)$.

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I interpret your question about group algebras to really be something like the following: What extra structure do group algebras have that allows you to "remember" that they come from groups, and in particular how do you see the group theory axioms in terms of this structure? The answer is that group algebras have the additional structure of a Hopf ...

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