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1

As Zhen Lin said in the comments, there is a very general argument that answers your problem. Denote $\omega$ for the category of finite ordinals with set-functions between them. Definition 1. A Lawvere theory is a finite-product-preserving bijective-on-objects functor $\ell \colon \omega^\circ \to \mathcal T$. A morphism $f$ from the Lawvere ...


1

Yes, as noted in comments, your argument works. If $K$ is a compact subset of $X$ and $U$ is open subset of $Y$, then the inclusion $A\subset Y$ maps $\{f\in C(X,A):f(K)\subset A\cap U\}$ to the $\{f\in C(X,A):f(K)\subset U\}\bigcap C(X,A)$. So the map $C(X,A)\to C(X,Y)$ identifies elements of subbase of own compact-open topology of $C(X,A)$ with elements ...


4

This process stabilizes imediately. To see this, note a somewhat more general result: if a diagram $D$ has an initial object $i$, then every functor $F : D → \mathscr C$ has a limit $Fi$ (with the obvious limiting cone). Given another cone $ΔC ⇒ F$, the component $C → Fi$ at $i$ is the required unique factorization through the limiting cone. In your case, ...


2

If $F,G:\mathcal A\rightarrow\mathcal B$ are functors and $\alpha:F\stackrel{\bullet}{\rightarrow}G$ is a natural transformation such that $\alpha_{A}:F\left(A\right)\rightarrow G\left(A\right)$ is invertible for each object $A\in\mathcal A$ then it can be shown that the inverses $\alpha_{A}^{-1}:G\left(A\right)\rightarrow F\left(A\right)$ are the components ...


3

Not necessary, consider the directed graph on $A,B,C,B',C'$ with arcs $A\to B$, $B\to C$, $A\to C$, $A\to B'$, $B'\to C'$, and $A\to C'$. This directed graph is a counterexample to your claim since $A\to B$ and $A\to B'$ do not lead to a common vertex. P.S. Combinatoric is not a word, combinatorial is.


3

First of all, a diagram in $\mathcal{C}$ is not the same thing as a directed graph. A $\Gamma$-shaped diagram in $\mathcal{C}$ is a morphism of graphs $\Gamma\longrightarrow U(\mathcal{C})$, where $U(\mathcal{C})$ denotes the underlying directed graph of $\mathcal{C}$ ${}^{1)}$. Secondly, it makes sense to talk about commutativity for all kinds of graphs. ...


1

The category $(A\downarrow \mathcal{M}\downarrow B)$ is not necessarily a model category, as it can fail to be complete or cocomplete (or both). For example, $(\{*\} \downarrow \mathsf{Set} \downarrow \varnothing)$ is not a model category: it is empty! There is no map from a singleton to the empty set. However one can define, for every $f : A \to B$, the ...


9

0) The excellent mathematician you evoke has as family name (=surname) tom Dieck and as first name Tammo: tom is part of his surname and has nothing to do with Tom, the endearing form of Thomas. 1) Your idea of "finding a book that starts with a formal treatment of the basics of category theory and moves to more advanced/specialized concepts in a ...


1

In Functional Analysis, one often speaks of projective limits (limits in the category LCS of locally convex spaces) and inductive limits (colimits in LCS). In the latter case one has to be careful as colimits in TOP usually differ from those in LCS. What you state as a theorem says that every locally convex space is a projective limit of seminormed spaces. ...


5

The pullback of two morphisms $f: A\to B$ and $g: C\to B$ in a category ${\mathscr C}$ is, if it exists, the product in the overcategory (or slice category) ${\mathscr C}/B$, the latter being the category whose objects are morphisms $X\to B$ in ${\mathscr C}$, and whose morphisms $(X\to B)\to (Y\to B)$ are the morphisms $X\to Y$ in ${\mathscr C}$ such that ...


0

Concerning 1, yes. This is an example where abstract nonsense is the most useful: An inverse limit is a categorical limit in the category of $C^*$-algebras, and those are unique up to unique isomorphism. If you check that $A_0$ fulfills the universal property, that means automatically that $A_0 \cong \lim{A_I}$. A proof goes like that: Assume there's a ...


5

Yes. This is a corollary of Schur-Weyl duality. You need at least the additional assumption that your symmetric monoidal category is enriched over $k$-vector spaces. In general I don't see any reason to expect that the action of $k[S_n]$ is faithful; consider, for example, the special case where we only look at $1$-dimensional vector spaces. Sometimes. The ...


0

Here's what I tried to do. Let's construct a homomorphism $F(A) \xrightarrow{\rho} F(A\cup A')/R''$, where $R''$ is the normal closure of $\Psi \cup \Psi'$. Define $\rho(w) = wR''$. Then by universal property of quotients(and since the normal closure of $\Psi$ $R \subseteq R''$) we have a unique homomorphism $F(A)/R \xrightarrow{i_1} F(A \cup A')/R''$ ...


3

Let $j : U \to X$ be the inclusion of an open subspace. Then $j^* : \mathbf{Sh} (X) \to \mathbf{Sh} (U)$ has a left adjoint $j_! : \mathbf{Sh} (U) \to \mathbf{Sh} (X)$: given a sheaf $F$ on $U$, $$j_! F (V) = \begin{cases} F (V) & \text{if } V \subseteq U \\ \emptyset & \text{otherwise} \end{cases}$$ the idea being that $(j_! F)_x = F_x$ for $x \in ...


8

The category of left $G$-sets is equivalent to the category of left $H$-sets if and only if $G$ is isomorphic to $H$. Indeed, consider $G$ as a left $G$-set. It has the property that $\mathrm{Hom}_G (G, -) : G \textbf{-Set} \to \mathbf{Set}$ preserves all colimits. Moreover, up to isomorphism, $G$ is the unique such left $G$-set, i.e. if $X$ is a left ...


0

If you consider the study of separable metric spaces a valid field, that provides an example (although by the wrong reason): They have cardinality bounded by that of $\mathbb{R}$.


1

To elaborate on Qiaochu's comment: The theorem written on the nLab states that $\mathcal G$ is separable iff the counit $\epsilon$ has a retraction, i.e. a left inverse. This implies that $\epsilon$ is a monomorphism, but is not equivalent to it. In case that was your confusion: Qiaochu himself has written a nice post on his blog that I read a few weekss ...


1

A map out of a group with presentation $(A,\Psi)$ is just a choice of image for each element of $A$ that respects the relations $\Psi$. So a map out of $(A\sqcup A', \Psi\sqcup \Psi')$ is just a choice of images for each element of $A\sqcup A'$ respecting the relations in $\Psi$ and in $\Psi'$. But since the relations in $\Psi$ have nothing to do with $A'$ ...


0

Here's a simple case when $A$ and $A'$ aren't disjoint: let $A = A' = \{ x \}$, and let $\Psi = \Psi' = \varnothing$. Then $G=G'=F1$, that is, $G$ and $G'$ are both the free group on one generator. The coproduct of $G$ and $G'$ in $\mathsf{Grp}$ is the free group on two generators, however $$(A \cup A' \mid \Psi \cup \Psi') = (\{x\} \mid \varnothing) = ...


3

Yes, this is a covariant functor $Set\times\mathcal{C}\to \mathcal{C}$. Given $f:S\to T$ and $g:C\to D$, the induced map $(f\cdot g):S\cdot C\to T\cdot D$ is the unique map such that if $s\in S$ and $i_s:C\to S\cdot C$ is the corresponding inclusion, then $(f\cdot g)i_s=i_{f(s)}g$, where $i_{f(s)}:D\to T\cdot D$ is the inclusion corresponding to $f(s)\in ...


6

The paper Seven Trees in One exhibits a "very explicit bijection" $T^7\cong T$. It is perhaps a bit cumbersome because it requires separating into five cases based on how the seven trees look in the first four levels of depth. A proof is present too. Disclaimer: I haven't read it. Another paper Objects of Categories as Complex Numbers discusses the ...


1

Let me first say that it is kind of strange to start with the total space of a trivial bundle $E = M \times F$, which implies that all fibers are equinumerous, and then write "Due to the inhomogenity of the fibers (i.e. the cardinality and the structure of the set $Σ_x$ and therefore morphisms between this set is not independent on $x∈M$) [...]", which ...


1

Products in Set are Cartesian products. Coproducts in Set are disjoint unions. What is said about the category of sets concerning Cartesian products and disjoint unions must be stated and proved by you in general for categories that have binary products and binary coproducts respectively. Then concerning products and coproducts. It is not a matter of "using ...


3

Take $R=\mathbb{Z}$, $M=\mathbb{Q}\oplus \mathbb{Z}/2\mathbb{Z}$ and $N= \mathbb{Z}/2\mathbb{Z}$, with inclusion $R\to M:n\mapsto (n,\overline{0})$. Then, in $M\otimes_\mathbb{Z} N$, we have that $$ (1,\overline{0})\otimes \overline{1} = (\frac{1}{2},\overline{0})\otimes 2\cdot\overline{1} = (\frac{1}{2},\overline{0})\otimes \overline{0} = 0.$$ Therefore ...


4

First of, a functor $F : \mathscr C → \mathrm{Set}$ is called representable (by $C$) if it's isomorphic (not necessarily equal) to $\mathrm{Hom}(C, -)$ for an object $C$ of $\mathscr C$. As for the term, an abstract functor $F$ is represented by the very concrete action of $\mathrm{Hom}(C, -)$. Take for example the functor $L : \mathrm{Top} → \mathrm{Set}$ ...


1

"Superfluous epimorphism" is dual to "essential monomorphism": An epimorphism $f: X \rightarrow Y$ is superfluous if for any morphism $g: T \rightarrow X$ one has $f \circ g \text{ epi} \Rightarrow g \text{ epi}$. In the case of modules this is equivalent to $\ker(f)$ being a superfluous submodule of $X$. In one direction, observe that $f\circ g$ epi ...


3

Let $(F, G, η, ε)$ and $(F', G', η', ε')$ be adjunctions, $F: \mathscr C → \mathscr D$, and $F' : \mathscr C' → \mathscr D'$. There is a powerful isomorphism of functors $\mathrm{Nat}(F'-, -F)$ and $$\mathrm{Nat}(-G, G'-) : [\mathscr C, \mathscr C']^{\mathrm{op}} × [\mathscr D, \mathscr D'] → \mathrm{Set},$$ and in particular, for every $K : \mathscr C → ...


7

Let $n \geq 2$. If $A,B$ have a product $P$ in the category of sets with $n$ elements, then $\hom(A,P) \cong \hom(A,A) \times \hom(A,B)$ shows $n^n = n^n \cdot n^n$, a contradiction.


4

Product of two objects in category is unique up to isomorphism. That is, if you found two objects that satisfy the definition of product, then there exists a unique isomorphism between these two products that respects the product structure (remember that a product is not just an object, but also two projection morphisms). This allows one to speak about the ...


3

One way to approach this is to start with $\in $, and build up a "universe" $U$, using a standard naive set theoretic construction. $U$ is defined in such a way so as to ensure that all the usual operations of set theory applied to elements of $U$, always produce elements of $U$. Now, we say that $u$ is $\textit {small}\Leftrightarrow u\in U$. Otherwise ...


6

The notion of a class is defined rigorously in Von Neumann–Bernays–Gödel set theory, which is a conservative extension of ZFC. Basically, you can form classes of sets using unrestricted comprehension, and you can freely take subclasses, images of classes under functions, and use the Axiom of Choice on classes of sets. However, no proper class is allowed to ...


6

It is just that, a class. That is: Given $A$, it is either true or false that $A\in\operatorname{Obj}(\mathcal C)$. Put differently, we may assume that "is an object of category $\mathcal C$" is a valid predicate.


1

Don't take my answer as a reference, but from my experience, notably in French, "action of $G$ on $X$" usually refers to a set-action (i.e. an homomorphism $G \to \operatorname{Aut}_{\mathbf{Set}}(X)$) while "representation of $G$ on/in $X$" refers to an homomorphism $G \to \operatorname{Aut}_{\operatorname{Mod}(T)}(X)$. For example, one talks about "linear ...


2

Here is a way to see it. Let $\mathcal C $ be a category with finite products. By Yoneda, the functor $\mathfrak h \colon \mathcal C \to \widehat{\mathcal C}, c \mapsto \hom(-,c)$ is fully faithful. Moreover, it preserves finite products by definition. So Lemma 1. An object $g$ in $\mathcal C$ is a group object with multiplication $m$, inverse $i$ and ...


2

$C^{2}$ has objects that are arrows $f:a\to b$ of $C$. A morphism $\phi :f\to g$ is a pair $(h,k)$ such that $k\circ f=g\circ h$ $\tag 1 \begin{matrix} \operatorname a & \xrightarrow{{f}} & \operatorname b \\ \left\downarrow h\vphantom{\int}\right. & & \left\downarrow k\vphantom{\int}\right.\\ \operatorname c& \xrightarrow{g} & ...


3

Your mistake is trying to think of morphisms as transformations; morphisms in this category are simply commutative squares; they don't have to "do" anything, aside from satisfy the algebraic properties of a category. Also, the pair $(h,k)$ is not necessarily sufficient to uniquely identify a morphism in this category: you may need more. e.g. the additional ...


1

As you point out, if the sets $A$ and $B$ are abelian groups, then the product and coproduct are the same. This means that the object $A\times B\in AGrp$ is a $\textit coproduct$. and we have the canonical insertions $$i_{1}:A\to A\times B;a\mapsto (a,1)$$ and $$i_{2}:B\to A\times B;b\mapsto (1,b)$$ that satisfy the UMP of the coproduct. We will show by ...


4

At least in algebraic contexts, I like to think of the coproduct of $(A,\cdot)$ and $(B,*)$ as being the structure which $A\cup B$ generates* (applying the original multiplication between pairs of adjacent elements of $A$ or of $B$). That is, every element of the coproduct must be writable as a product of $A$'s and $B$'s - like (assuming associativity): ...


0

One possibility (following Takeuchi) is as follows: Let $C$ be a coalgebra over a field $k$ . The free Hopf algebra $(H(C), i)$ generated by $C$ is characterized by the following universal property: (1) $i:C\rightarrow H(C)$ is a coalgebra map (2) $Hom(i, H)$ : Hopf $(H(C), H)\rightarrow Coalg(C, H)$ is a bijection for any Hopf algebra $H$.


3

Commutativity has to do with how much a monoid object can be delooped. In general, there are infinitely many levels of delooping corresponding to infinitely many levels of commutativity, and monoids which can be delooped $n$ times are called $E_n$ algebras. One way of saying what this means is that an $E_n$ algebra in a (symmetric monoidal, or $E_{\infty}$) ...


3

Commutative rings are monoidal $\mathsf{Ab}$-categories with one object. (Yes, symmetric monoidal is not assumed; it follows.)


4

You have $\text{Hom}(A,G(-))=\text{Hom}(A,-)\circ G$, so you need exactness of both $\text{Hom}(A,-)$ (i.e., projectivity of $A$) and $G$ to ensure exactness of the composition.


1

In mathematics, when you study sets with some extra structure/operation (for ex. multiplication in groups), you are also interested in studying functions that "respect" that structure (for ex. group-homomorphisms). With presets (also known as prosets=PRe-Ordered SETS) the structure/operation is the pre-order $\le$. So with groups you are interested in ...


1

The category of $C^*$-algebras is complete like Martin Brandenburg explained. However, given an inverse system of $C^*$-algebras, it is often more convenient to consider its limit in the category of all topological $*$-algebras. The limit of a diagram $(A_i,\lVert - \rVert)_{i \in I}$ of $C^*$-algebras is then the topological $*$-subalgebra of the ...


1

Yes, they can be different in nature: Consider e.g. the identity map for $f$ from $R=(\Bbb N,\,|\,)$ to $S=(\Bbb N,\le)$.


2

$Z_n\left(M\right)$ is the $n$-th homogeneous component of the $0$-th cyclic homology of the tensor algebra $T\left(M\right)$. There is a nice expository note about this by Clas Löfwall, including the relation to the logarithm: Clas Löfwall, Cyklisk homologi, 28 Aug 2012. Despite the title, it is in English.


1

As the initial comments on the question make clear, the first response to the question is "It's a one-line proof by duality!". For the result that a category that has all coproducts and coequalizers is cocomplete is of course an immediate corollary by duality of the standard result that a category that has all products and equalizers is complete. So is ...


1

The naturality squares of a natural transformation are: $$\require{AMScd} \begin{CD} FA @>{\eta_A}>> GA\\ @V{Ff}VV @VV{Gf}V\\ FB @>>{\eta_B}> GB \end{CD}$$ where we ask the square to commute for any choice of $f\in \mathsf{Hom}(A,B)$. To me, the significance of this definition is somewhat philosophical (and has hitherto given me good ...


3

This is getting too long for a comment, but here are some more examples: Modules over a division ring $K$ work. Also, pointed sets work. Affine spaces over a field, or more generally a division ring, work. These examples come from here where Lawvere theories with all algebras free are considered. Some observations from over there still apply. Your ...


2

You may be trying to be a bit too formal. Here's how I'd explain it. $\lambda$ is a map out of a colimit, so it's got components $\lambda_C$ for every $C$. $\lambda_C$ is a map into a limit, so it's got components $\lambda_C^D$ for every $D$. $\lambda_C^D:\varprojlim_D F(C,D)\to \varinjlim_C F(C,D)$ is given by projecting to $F(C,D)$ from the limit then ...



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