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1

The difference between the definitions is explained in the first paragraph of section 2: their definition, unlike yours, allows for multiple edges with the same vertex set. So, for a silly example, if $X$ is a set, $x∈X$, and $G_i$ are a bunch of groups acting on $X$, then we could view each group $G_i$ as an edge whose vertex set is the $G_i$-orbit of $x$. ...


0

Let $\mathcal{C} $ be a $2$-category and $A$ be an object of $\mathcal{C} $. In most of the cases, there are lax natural transformations $\mathcal{C}(A,-)\to\mathcal{C}(A,-) $ which are not $2$-natural or even pseudonatural. For instance, take the $2$-category generated by two parallel arrows $f,g: A\to B $ with a $2$-cell $f\to g$. We have that ...


0

Mac Lane showed that any combination of multiplications lead to the same function, if they have the same domain and codomain, via coherence. Since both $F(g \circ h)$ and $F(g) \circ F(h)$ are combinations of multiplications, they are equal via that theorem.


2

I'll explain how to complete the proof you started: If $x=(a,t)∈U$, the goal is to find an open product $W×V⊆U$ such that $W$ is saturated, as that would imply $(φ×1_T)(W×V)=φ(W)×V$ is an open neighborhood of $(φ(a),t)$ contained in $(φ×1)(U)$. So take a compact neighborhood $K$ of $t$ such that $\{a\}×K⊆U$, and let $W$ be the largest set such that $W×K⊆U$. ...


1

No. A counter-example is the Yoneda embedding: take $A=B=Ab$ to be the category of abelian groups, and let $A'=Fun(A^{op}, Ab)$ be the category of additive contravariant functors from $A$ to $Ab$. Then $A'$ is an abelian category, and the Yoneda functor $$ h:A\to Fun(A^{op},Ab): X\mapsto Hom(-,X) $$ is fully faithful. Let $B'$ be the image of $h$. ...


1

There is a coproduct in the category of groups, namely the free product. Perhaps you mean the category of finite groups, which does not have a coproduct. Lang says that the coproduct in the category of groups is not the product; in any abelian category finite products coincide with finite coproducts, e.g., in the category of abelian groups.


0

You ask several questions, but I will only address the one in the title. No, this is in fact not the right way to think about groups as one-object all isos categories. While the details you give are correct, it is important to realise that the precise details of what the single object in the category is, is irrelevant. In fact, there is an equivalent ...


0

Let $(G, \circ)$ be a group. Let $\mathcal{C}$ be the subcategory of $Set$ consisting of the single object $G$, and $\mathcal{C}(G, G) := \{f_g: g \in G\} \subset Set(G, G)$, where $f_g$ is the function - i.e. morphism in $Set$ - satisfying $f_g(h) = h \circ g$ for all $h \in G$. Then we have $f_{g \circ g'} = f_g f_{g'}$, $f_e = 1_G$, and $f_g$ is an ...


3

Pretend that $A,B,C$ are sets, and that $Fun$ stands for functions rather than functors. Can you show an isomorhpism between the set $Fun(A\times B, C)$ and $Fun(A, Fun(B,C))$? Hint: if $f\colon A\times B \to C$ is a function, and now you fix some value $a_0\in A$, can you construct in a natural way a function $g_{a_0}\colon B\to C$? Once you answer this ...


2

Often it is much more convenient to work with representable functors. The universal property of the group $(A|R)$ is the following: If $H$ is a group, there is a natural isomorphism (i.e. bijection) $$\hom((A|R),H) \cong \{f \in \hom(A,|H|) : f(R)=\{1\}\},$$ where by $f(R)$ I mean the set of all $f(r)$, $r \in R$, and $f(r)$ is defined by extension of $f$ ...


1

No, you don't need the maps to be surjective. The category of groups is complete (all limits exist) so you don't need any requirements on the morphisms here. You can easily just construct the fibre product by taking the product of the sources $G\times H$ and then the equalizer of $f \circ \pi_1$ and $g \circ \pi_2$.


3

Yes, see here. A standard reference is Kelly's Basic concepts of enriched category theory, TAC reprint. The case of $k$-linear categories is very simple (basically because here the monoidal unit is a generator). Here, $k$ can be any commutative ring. For $k$-linear functors $F : \mathcal{C} \to \mathsf{Mod}_k$ and objects $A \in \mathrm{Ob}(\mathcal{C})$ ...


1

So you have arrows $f:I\to[A,A']$ and $g:I\to[B,B']$. The functor we are interested in and which you called $S$ is the composite $$ \mathcal V_0\times\mathcal V_0\to (\mathcal{V\otimes V})_0\to \mathcal V_0 \\ \mathcal V_0(I,[A,A'])×\mathcal V_0(I.[B,B'])\to \mathcal V_0(I,[A,A']⊗[B,B'])\to \mathcal V_0(I,[A⊗B,A'⊗B']) \\ (f,g)\mapsto (f⊗g)l^{-1}\mapsto ...


1

If we don't have the surjectivity we will loose Goursat's lemma. Since a fiber product will no longer be always a subdirect product.


2

I guess by $(R)$ you mean the normal closure $\langle R^{F(A)} \rangle$ of $R$ in $F(A)$? There is a unique homomorphism $\psi:F(A \cup A') \to H$ with $\psi(a) = f(a)$ for $a \in A$ and $\psi(a) = g(a)$ for $a \in A'$. Since all elements of $R$ lie in $\ker(f)$ and all elements of $R'$ lie in $\ker(g)$, we have $R \cup R' \subset \ker(\psi)$ and hence, ...


2

A preorder is essentially a category $\cal P$ whose hom-sets $\mathcal P(x,y)$ have at most one arrow $x\to y$, which then represents the relation $x\le y$, otherwise the hom-set is empty. The coproduct of a family of objects $(x_i)_i$ is an object $s$ such that $x_i\le s$, and the universal property is just $s$ being the least upper bound, i.e. whenever ...


4

A group is essentially the same thing as a category satisfying the following two conditions: 1) there is only one object; and 2) every morphism is an isomorphism. Of course not every category has these two properties, so there are many more categories than groups (e.g., the category of sets, the category of categories, the category of groups, etc.). The ...


4

So suppose $$\begin{array} AA & \stackrel{g}{\longrightarrow} & C \\ \downarrow{f} & & \downarrow{\beta} \\ B & \stackrel{\alpha}{\longrightarrow} & D \end{array}$$ is a pushout and $f$ is epi. We will show that $\beta$ is epi. Suppose, $h_1, h_2 \colon D \to D'$ are such that $h_1 \beta = h_2 \beta$. We have $$ h_1\alpha f = h_1 ...


7

The duality is that between extremal monomorphisms and extremal epimorphisms. A monomorphism is the right generalization of injective function to situations with more structure, such as topological spaces with continuous functions. We say $f:X\to Y$ is a monomorphism if, whenever $g,h:Z\to X$ are such that $f\circ g=f\circ h$, we can infer $g=h$. ...


1

Actually, it might be that Mac Lane made a small error (gasp!) This error can be seen simply by seeing the domain and codomain don't match up. $$\mu^{k_1 + k_2 + ... + k_n}:c \otimes c \otimes \dots \otimes c \space (k_1+\dots+k_n \text{times}) \to c$$ $$\mu(\mu^{k_1} \otimes \mu^{k_2} \otimes \dots \otimes \mu^{k_n}):(c \otimes \dots \otimes c \space (k_1 ...


1

Let $\varphi:S\to T$ be your homomorphism. We need to check that $(T,\cdot)$ satisfies the axioms of a group. If $e$ is the identity of $S$, we need to show that $\varphi(e)$ is the identity of $T$. Since $\varphi(a)\varphi(a^{-1})=\varphi(e)$, we can show that every element of $T$ has an inverse ($\varphi$ is surjective). Associativity follows from the ...


4

The only hypothesis we are given is that these are two structures with binary operations. The fact that there is a homomorphism between them a priori implies nothing without knowing something about the homomorphism. Some examples: There is a homomorphism from $\mathbb{Z}_4$ into the multiplicative monoid of the Gaussian integers $\mathbb{Z}[i]$ sending a ...


4

This is false. Taking these two facts: tensoring $- ⊗_R M$ is exact iff M is a flat module a module over a PID is flat iff it is torsion free we see that $- ⊗ \mathbb Q : \mathrm{Ab} → \mathrm{Ab}$ is an exact functor, but it doesn't preserve free objects: $\mathbb Z ⊗ \mathbb Q = \mathbb Q$, and $\mathbb Q$ is not free over $\mathbb Z$ (any two elements ...


1

As pointed out in the comments, there is an error in the nlab article so in fact the source category for the homology functors should not be $\mathsf{Top} \times \mathsf{Top}$. Instead, the source should be the subcategory whose objects are embeddings $\iota_A: A \hookrightarrow X$, and whose morphisms from $\iota_A: A \hookrightarrow X$ to $\iota_B: B ...


4

I suggest Peter May's Concise course on algebraic topology. You will find e.g. categorical formulations (and proofs) of the van Kampen theorem and the classification of covering spaces.


4

The closest thing I've found is Strom's Modern Classical Homotopy Theory, although I haven't read much of it. Chapter 1 is called Categories and Functors, so that's a good start. This is the only introductory algebraic topology textbook I know of that explicitly uses the language of homotopy limits and colimits.


2

Rotman's An Introduction To Algebraic Topology is a great book that treats the subject from a categorical point of view. Even just browsing the table of contents makes this clear: Chapter 0 begins with a brief review of categories and functors. Natural transformations appear in Chapter 9, followed by group and cogroup objects in Chapter 11. The aspect I ...


4

Spanier's book is relatively old (so I know it does not quite answer your question), but excellent. It uses category theory from the get-go. Riehl's "Categorical homotopy theory" is very well-written, though it may be a bit too advanced if you hadn't seen a bit of algebraic topology already. Riehl's book is focused on the categorical aspect via Quillen model ...


0

Ronald Brown's text Topology and Groupoids is probably what you want from a topology text. He gives an introduction to general topology and the fundamental groupoid using the language of category theory throughout. It's an excellent textbook.


2

I guess (from the comment discussion mostly) your question is not really about free objects but rather: given an adjunction $F \dashv U$, how can I explicitly write the bijection $\hom(FA, B) \to \hom(A, UB)$ so that the special case $\mathsf{Set} \leftrightarrows \mathsf{Grp}$ gives me the restriction $\varphi \mapsto \varphi\restriction A$? Given $F ...


1

The straightforward answer to your question is: sure, but they're not very interesting. If we view a (2-valued) model $M\models ZF$ (I use $ZF$ for simplicity) as a category whose objects are sets in $M$ and whose morphisms are functions in $M$, as you do, then - whenever we have an end-extension $N\supseteq M$ - the inclusion map is a functor from $M$ to ...


3

You only need to understand what the first dual map $\phi^\vee$ is, as $(\phi^\vee)^\vee=\phi^{\vee\vee}$. The map $\phi^\vee$ is defined by $\phi^\vee(f)=f\circ\phi$. This is just a special case of the contravariant hom functor. To check that the diagram is commutative, notice that for $v\in V$ and $f\in W^\vee$ we have $$(\phi^{\vee\vee}\circ ...


3

Yes, absolutely. The approach I know about is synthetic differential geometry, which begins with thinking of manifolds not via their underlying locally Euclidean topology (this is analytic differential geometry!) but via their smooth functions, in particular those to and from the line $R$ ($R$ behaves differently enough from $\mathbb{R}$ that it's a good ...


6

If $A$ is any ring, then $A$ and $M_n(A)$ have equivalent categories of modules, and usually $A$ and $M_n(A)$ are not isomorphic. This is the simplest example of a Morita equivalence.


1

I like the picture of a cocone from http://arxiv.org/abs/math/0306223: They describe it through a story about people sending emails to each other. And then: A cocone which is universal is a colimit. ―http://ncatlab.org/nlab/show/cocone Added: Colimits glue. Limits cut. ―Adam Hughes


0

So let $G:C\to D$ be faithful functor, and assume $D$ has equalizers, and that $(G\downarrow x):(C↓x)\to(D↓Gx)$ has a right-adjoint-right-inverse $L$ (meaning that the counit is always the identity). If $f:f':x\to y$ are parallel arrows, we apply $G$, which gives us distinct arrows $Gf,Gf'$, with an equalizer $t:s\to Gx$. Applying $L$ to $t$ gives us ...


5

An explicit counterexample is $\text{Ab}^{op}$ (an excellent counterexample in general), which by Pontryagin duality is equivalent to the category of compact (Hausdorff) abelian groups. The product is what you think it is but the coproduct is more complicated: it is not the direct sum but the Bohr compactification of the direct sum, and because this ...


2

If $X\stackrel f\leftarrow A\stackrel g\to Y$ is a cotriad, the pushout is $$X\stackrel{\bar g}\to Z\stackrel{\bar f}\leftarrow Y$$ where $Z$ is the quotient space of $X\sqcup Y$ by the relation generated by $f(a)\sim g(a)$, and $\bar f$ is the composite $Y\to X\sqcup Y\to Z$. As often with relations, one only says what generates it, and the whole relation ...


1

First note that a monomorphism $f:X\to Y$ must be injective: If $f(x)=f(y)$, then the compositions of $f$ with the maps $(*\to X)$ sending the single point to $x$ and $y$, respectively, are equal, thus these maps are equal, hence $x=y$. So let $f:X\to Y$ be a monomorphism in $\mathbf{CHaus}$. Since a map from a compact space to a Hausdorff is perfect ...


0

Let $\mathbf C$ be a category and consider an adjunction $F\colon {\mathbf Set} \leftrightarrow \mathbf C\colon U$, with left adjoint $F$. Assume there exists some object $A$ in $\mathbf C$ with $|UA|\ge 2$. Then for all sets $S$ the unit $\eta \colon S\to UFS$ is injective. (The result you are interested in now follows by noting that monoids with more ...


1

In principle, "$F$ is a left adjoint" is fine because the adjunction is determined uniquely up to unique isomorphism by $F$. This seems to contradict your intuition that "$F$ is a left adjoint" sounds wrong, but it's perfectly consistent with standard usage such as talking about "the" limit of $G$: this sort of usage works fine except in contexts where it's ...


8

In general, in a category $\mathsf{C}$ that has products, an object $c$ such that $- \times c : \mathsf{C} \to \mathsf{C}$ has a right adjoint is called exponentiable. There are various characterizations of exponentiable objects in $\mathsf{Top}$, and you can find some of them in this $n$Lab article: Theorem (Exponentiability, I). An object $X$ of ...


2

There is no neighborhood filter in $\mathbb{Q}$ of $x\in\mathbb{R}\setminus\mathbb{Q}$, at least not one satisfying the correct axioms. In $\mathbb{Q}$, we have $(x-1,x+1) = (x-1,x)\cup(x,x+1)$. So the filter of open sets containing $x$ cannot be prime, and should be illegal in whatever source you're using.


5

$\mathbb{Q}$ is a Hausdorff space. Every Hausdorff space is sober, so its soberification is just itself.


5

You're working too hard. I'll assume you're convinced that $\mathbb{Z}$ is the free group on one generator, which means you're convinced that $$\text{Hom}(\mathbb{Z}, G) \cong G$$ (where the RHS should really be the underlying set of $G$ but I'm omitting that I'm applying the underlying set functor). One way to state the universal property of the coproduct ...


0

The set $X\times Y$ together with the projection $\pi_X:X\times Y\to X$ and $\pi_Y:X\times Y\to Y$ serves as the categorical product in the category of sets, that means, given a set $A$ and maps $f:A\to X$ and $g:A\to Y$, there is a unique map $\phi=(f,g)^\flat:A\to X\times Y$ such that $\pi_X\phi=f$ and $\pi_Y\phi=g$. It's clear that this map must be ...


2

If you want to write a map $S : X \times Y \to Y \times X$, you could use the UPP for $Y$ and $X$: given two morphsims $f : A \to Y$ and $g : A \to X$, there exists an unique morphism $h = \langle f , g \rangle : A \to Y \times X$ such that $\pi_Y \circ h = f$ and $\pi_X \circ h = g$. To find $S$ just take $A = X \times Y$, $f = \pi_Y$ and $g = \pi_X$, and ...


1

No. Consider the functor $\Delta : \mathbf{Set} \to \mathbf{sSet}$ that sends each set $X$ to the discrete simplicial set on $X$. This is certainly fully faithful, and it has a left adjoint, namely the functor $\pi_0 : \mathbf{sSet} \to \mathbf{Set}$ that sends a simplicial set to its set of connected components. It is not hard to check that $\pi_0 : ...


7

An endomorphism is a morphism from an object to itself, full stop. I don't know why you're so distrustful of people who aren't category theorists (I'm a graduate student who studies category theory myself; is that good enough or do I need a PhD first?), but, for example, this is the definition you will find on Wikipedia, the nLab, a different page of ...


6

Surjectivity has nothing to do with being endomorphism. Think for example a constant map from a set of more elements to itself: that's a not surjective endomorphism in the category of sets and functions.



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