# Tag Info

2

Your "construction" works just fine for the following reasons. If $f^{-1}x$ is not empty then $gx$ is independent of the choice $f'x \in f^{-1}x$ by the assumption $fa_1 = fa_2 \Rightarrow ha_1 = ha_2$ for $a_1,a_2 \in A$. For $a \in A$ we have $f(f'f(a)) = f(a)$ by the definition of $f'$, so $gf(a) = hf' f(a) = h(a)$ by the assumption. The question is ...

3

Edit: This answer is basically the same as Baby Dragon's. I didn't see his/her answer when I started working on my pictures! Let $A=\{A_i:i\in I\}$ and $\{B_j:j\in J\}$ be two collections of $\mathscr C$-objects such that $\coprod A$ exists and $\prod B$ exists. For each $i\in I$ and $j\in J$ let $\ell_{ij}:A_i\to B_j$. For fixed $j\in J$, the universal ...

3

Let me explain something more general. Let $\{X_i\}$ and $\{Y_j\}$ be two collections of objects. In order to specify a map $$f:\coprod X_i\to \prod Y_j,$$ it suffices to say how we map $\coprod X_i$ to each factor, $Y_j$. Thus we need to define a set of maps $$f_{\bullet j}\coprod X_i\to Y_j.$$ But to specify a map out of a coproduct, it suffices to give a ...

1

Your definition of logic is pretty much correct. A logic contains both the language which the signature $\Sigma$ generates and the deductive system defined by $\vdash$. A type theory is a logic with different sorts of individuals (called "types") and constructions that generate new types form existing ones, like product and arrow types. An internal logic ...

2

The right way to build such a category is a philosophical question. There are different approaches in the mathematical literature. One thing is clear though: the objects should be propositions, not just theorems. The problem is to define equality of proofs in a sensible way. For example, let $\Pi$ be Pythagoras' theorem. Should each of the over 100 proofs ...

3

See for example Lambek and Scott: Introduction to higher order categorical logic, ch 0.1 (unfortunately not in the net but for sure in ur university library). There first is defined a graph, then a deductive system as a graph with For each object $A$ an identity arrow $1_A:A\rightarrow A$ For each pair of arrows $f:A\rightarrow B$ and $g:B\rightarrow C$ ...

2

I don't have CWM here right now but ACC gives you an exact definition for the opposite category, and that answers your question at least partially. Here is the definition in ACC: For any category $\mathbf{A}=(\mathcal{O},\hom_\mathbf{A},id,\circ)$ the dual category of $\mathbf{A}$ is the category ...

4

If your maps are cofibrations, then the direct limit and the homotopy direct limit coincide (up to homotopy equivalence). So to find a counterexample, we must look at some less nice sequences of maps. For instance, consider the map $f : S^1 \to S^1$ defined by $e^{\theta i} \mapsto e^{2 \theta i}$. This is not a cofibration, obviously. The direct limit of ...

6

To clarify a point that may be confusing, $\mathbf{Rel}$ is one of the relatively rare categories equivalent to its own opposite, essentially by the isomorphism $A\times B\cong B\times A$. That's why you find a natural representative for a morphism in $\mathbf{Rel}^{op}$: you're actually demonstrating this equivalence. In a general category, there wouldn't ...

2

Strict equality is something that category theory really don't (shoudn't?) care about. The two opposite categories you construct are equivalent, and it is all that matters !

0

Let $X_n=\{0\}\cup[\frac1n,1]$ and $X_n\to X_{n+1}$ the obvious injection. Then $\mathop{\mathrm{dirlim}} X_n=[0,1]$ whereas the homotopy direct limit of $X_n$ is disconnected.

5

This paper, called Physics, Topology, Logic and Computation: A Rosetta Stone does just that in section 3.2. If you have time and interest, I would suggest reading the entire paper (since the whole thing is pretty cool).

5

False. A Cartesian Closed Category is an example of a Closed Category, and an example of this is $\mathbf{Set}$, where the internal hom functor corresponds to the exponential functor, so $\mathrm{hom}(X,Y)=Y^X$. Then the LHS of the equation becomes $Z^{(Y^X)}$ while the RHS is $(Z^X)^Y$. Whether you're asking for equality or just isomorphism doesn't ...

2

I think usually people denote it by $Mor(\mathscr{C})$. You should be careful, because it is not always a set (if the category $\mathscr{C}$ is big) as Alex G. pointed out in his comment. But this union of all morphisms is always again a category with object being morphisms in $\mathscr{C}$ and maps between morphisms being the obvious commutative squares.

11

A category consists of objects and arrows between them (which can be associatively composed). We can express various constructions and theorems with the help of categories, and when we switch the direction of all arrows, we arrive to the dual constructions and dual theorems, which are usually named by the prefix 'co', and which stay valid, as the ...

1

The internal logic of a topos is an instance of the internal logic of a category (since toposes are special kinds of categories). The internal logic of toposes (instead of an arbitrary category) can also be interpreted with the Kripke-Joyal semantics. For more on this, check part D of Johnstone's Elephant and chapter VI of Mac Lane's and Moerdijk's Sheaves ...

2

Finite direct sums are preserved since we deal with additive functors. Now use the fact that direct sums are directed colimits of finite direct sums. Explicitly, we have $$\bigoplus_{i \in I} M_i = \operatorname{colim}_{E \subseteq I \text{ finite}}\, \bigoplus_{i \in E} M_i.$$

2

A category-valued sheaf transports a covering to a limit. A stack transports a covering to a 2-limit. As you say, the main difference is that a stack uses the 2-categorical structure of the 2-category of small categories.

2

Well the statement of the problem is not extremely clear, perhaps on purpose, but what it is trying to teach is the following: in the "example above" you conclude - by looking at the diagram without the $h$- that the composition of $f$ and $g$ gives an identity (on $A$ and on $B$). If they now tell you to add another - different - morphism $h$, you must be ...

2

$f$ and $h$ are distinct simply because the diagram says so. Any argument a la "They must be the same beacuase they are the unique twosided inverse of the isomorphism $g$" already assumes that we have a category.

2

Ittay Weiss: no need for symmetry at all. Zhen Lin: It is unusual to consider categories enriched in something other than a symmetric monoidal category in the first place – partly because a lot of the theory becomes awkward (e.g. opposite categories).

4

No, since $\hom(X,Y)$ is a set, but $Y^X$ is an object of $C$. The relation between these two guys is given by the following formula (where $1$ denotes a terminal object): $$\hom(1,Y^X) \cong \hom(X,Y).$$ This bijection follows immediately from the definitions. In other words, $Y^X$ is an object whose set of "global sections" is $\hom(X,Y)$. This is true ...

2

Take any abelian category $\mathscr A$ with insufficiently many projectives, and let $\mathscr F$ be the zero functor. Flat modules are precisely those which are acyclic for Tor (in either variable). In fact, $M$ is flat if and only if the functor $\text{Tor}_1(-, M)$ vanishes, if and only if the functor $\text{Tor}_i(-, M)$ vanishes for every $i>0$. As ...

5

There is a useful trick for dealing with this. The following appears as Lemma 1.3 in [Johnstone and Moerdijk, Local maps of toposes]. Proposition. Given an adjunction $$L \dashv R : \mathcal{D} \to \mathcal{C}$$ if $\mathrm{id}_{\mathcal{C}} \cong R L$ (as functors) then the unit $\eta : \mathrm{id}_{\mathcal{C}} \Rightarrow R L$ is (also) a natural ...

1

It means that $B \times C$ is terminal among categories equipped with projections on $B$ and $C$: for every category $D$ and every pair of functors $D \to B$, $D \to C$, there exists a unique function $D \to B \times C$ such that the diagram you posted commutes. For more information..

1

"Universal" here means that it is "terminal", in the sense that any other triple $(\mathbf D, R, T)$ (I employ your notations) admits a unique morphism, in a suitable category (exercise: define that "suitable category"), to $(\mathbf B \times \mathbf C, P, Q)$. This is precisely the universal property of the product, and defines it uniquely up to a unique ...

2

Let's start by labeling the arrows:$\require{AMScd}$ $$\begin{CD} A @>f>> B @>g>> C\\ @VVaV @VVbV @VVcV\\ D @>>h> E @>>k> F \end{CD}$$ The square $(X+Y)$ is exact if and only if the "diagonal sequence" $0 \to A \xrightarrow{(gf, a)^t} C \oplus D \xrightarrow{(-c,kh)} F \to 0$ is exact. The squares $(X)$ and $(Y)$ are ...

2

I'm not entirely sure that you want to read a category theory book. Category theory can be quite technical and dry, and difficult to understand if you don't have the motivating examples in mind. Moreover, it's a large subject and you'll only need the tools related to the part of knot theory you want to study. A better idea is to read a short expository ...

4

If $C$ is any category with a terminal object $t$ and $I$ is a category, then it is an easy exercise ($\approx$ 2 lines) that $\Delta(t)$ (the constant functor with value $t$) is a terminal object in the functor category $C^I$. Any other terminal object of $C^I$ is isomorphic to this one. Now apply this to $I=\{0 < 1\}$.

0

Regarding your second question: if $M$ is a bimodule, the complex $B(R,M)$ is indeed a complex of $R$-bimodules (the objects are bimodules and the maps are bimodule maps), and it is exact, but in general it is not projective. Indeed, $R\otimes M$ is not usually projective. Notice, in fact, that there are isomorphisms $\hom_{R{-}R}(R\otimes ... 0 I will try to answer to the first part of your question by introducing the Hochschild homology, showing how the bar resolution can be used to compute a$\operatorname{Tor}$of interest and relate it to the Hochschild homology. Let$R$be a$k$-algebra and$M$be an$R-R$bimodule. With$k$we denote a field and we write$\otimes$for$\otimes_k$. The ... 0 Héhé, I found this on internet, just a small paragraph on page 4 (or 246) talking about which norm one should take. 1 Hint. Take a morphism$f \colon 2 \to P$in$\mathsf{Pos}$such that$fu=fv$. What does it tell you about the image by$f$of the top and bottom of$2$? 4 By the universal property of being "free", there is at most one homomorphism$f\colon F \to F$, such that$i \circ f = i$(any set map$X \to H$into an arbitrary group [here$H = F$] extends to an unique homomorphism). You have two maps with this property, namely${\rm id}_F$and${\rm i}_{G \subseteq F} \circ \phi$. So this two maps must be equal, hence ... 3 A proof analogous to the one for tensor products will establish the following more general result that you can find in Milnor and Stasheff's Characteristic Classes: Let$\mathcal{V}$be the category of finite dimensional real vector spaces and isomorphisms of such (not all linear transformations, just isomorphisms, for maximum generality). Let$T : ...

4

The simplicial complex you describe is called the Tits building for $GL_n(k)$. When $k=\mathbb{F}_p$ it is homotopy equivalent to a wedge of $p^{n(n-1)/2}$ spheres of dimension $n-2$. I think that theorem is due to Quillen, and am pretty sure that volume II of Benson's Representations and Cohomology contains a proof. I don't know what happens for other ...

1

Consider the algebra of upper triangular $2 \times 2$ matrices$$\left( \begin{array}{cc} * & * \\ 0 & * \end{array} \right)$$ As a module over itself it is sum of two indecomposable modules. Denote them as $V_1$ and $V_2$ (one and two dimensional respectively). $P = V_1^{\oplus n} \oplus V_2 ^ {\oplus m}$ is a projective generator (here $n,m \geq ... 2 There is a useful trick for dealing with this. Proposition. Given an adjunction $$L \dashv R : \mathcal{D} \to \mathcal{C}$$ if$L R \cong \mathrm{id}_{\mathcal{D}}$(as functors) then the counit$\epsilon : L R \Rightarrow \mathrm{id}_{\mathcal{D}}$is (also) a natural isomorphism. Proof. Let$\delta = L \eta R$, where$\eta : \mathrm{id}_{\mathcal{C}} ...

2

Identity arrows are it. For an identity $x$ and functors $F,G$, the component of a natural transformation $\sigma:F\to G$ at $x$ is going to be a morphism for which $F(x)$ and $G(x)$ are left and right identities, respectively. Keep in mind in the normal definition of a category, there's a bijection between objects and their identities; indexing with ...

6

A natural transformation of functors $\mathbb{C} \to \mathbb{D}$ is the same thing as a functor $\mathbb{2} \times \mathbb{C} \to \mathbb{D}$, where $\mathbb{2} = \{ 0 \to 1 \}$. The domain of the natural transformation is the restriction to $\{ 0 \} \times \mathbb{C}$, and the codomain is the restriction to $\{ 1 \} \times \mathbb{D}$.

7

If $C$ is a closed monoidal category and $R$ is a monoid object in $C$, then a left $R$-module is the same as an object $M \in C$ together with a homomorphism of monoids $R \to \underline{\mathrm{End}}(M)$. If $R=\mathbf{1}_C$ is the initial monoid, it follows that every object of $C$ has a unique $R$-module structure. If $C$ is not closed, it can be also ...

3

Despite that these categories are equivalent, you have to be a little careful with the translation. For example, partial functions $X \times Y \rightarrow Z$ correspond to pointed-set homomorphisms $X \otimes Y \rightarrow Z$, where $\otimes$ is the smash product. This is because the free functor $F : \mathbf{Set} \rightarrow \mathbf{Set}_*$ satisfies the ...

1

I believe that all these usages of "elementary" refer back to the notion of an elementary class of structures. Let L be a language, and let $X$ be the class of L-structures. A subclass $\Sigma$ of $X$ is called elementary just in case there is a set $T$ of first-order axioms (in L) such that $\Sigma = \mathrm{Mod}(T)$.

8

It means that the categories are equivalent. Consider the functor $F : Pfn \to Set_*$ which maps a set $X$ to the pointed set $(X \sqcup \{*\},*)$, and a partial map $p : X \to Y$ to the pointed map $p_* : X \sqcup \{*\} \to Y \sqcup \{*\}$ which maps $* \mapsto *$, $x \mapsto *$ if $p(x)$ is not defined, and otherwise $x \mapsto p(x)$. It is easy to check ...

3

From the wikipedia page on simplicial sets, "Δ denotes the simplex category whose objects are finite strings of ordinal numbers..." I think that $\Delta\downarrow X$ is a slight abuse of notation, as the simplex category of a simplicial set $X$ is the same as the comma category $$\Delta^{(-)}\downarrow X$$ where ...

1

The quotient $G/[G,G]$ is known as the abelianization $G^{\rm ab}$. First check that $[G,G]$, the subgroup generated by the commutators, is indeed a normal subgroup: indeed, conjugation distributes over commutators as $[x,y]^a=[x^a,y^a]$. Next, check that $G\mapsto G^{\rm ab}$ is indeed functorial. Given a map between groups $\varphi:G\to H$, what is the ...

0

Here are some partial answers: A good definition of arrows for $\mathrm{Mod}(T)$ is elementary embeddings of models. It turns out that "being the category of models of a first-order theory" is not a property of a category. Let me explain: let C be the discrete category with an uncountable infinity of objects. Question: is C the category of models of a ...

0

Just to write down a plausible possibility. Copying from an answer in here one has (still a little bit formal) the coequalizer of $k$ and $l$ $$Coeq:= \coprod_{i\in I} F(U_i) \left. \right/ \sim$$ where $\sim$ is the finest equivalence relation such that $k(x)\sim l(x),\ \forall\ x\in\coprod_{i,j\in I}$ Still by imitating the answer, one would say that ...

2

In general, a $C$-valued precosheaf $F$ is a cosheaf iff for every object $T \in C$ the presheaf $\hom(F(-),T)$ is a sheaf. This should give you an intuition for cosheafs. Instead of talking about elements of $F(U)$, which are typically morphisms into $F(U)$, better think of "coelements", which might be morphisms on $F(U)$. If you want to consider elements ...

1

Let's do part of the task, to get you the feeling how to proceed with the other parts: $\emptyset$ is initial in the category of sets (and maps): If $A$ is any set, we need to exhibit a map $f\colon \emptyset\to A$ and show that it is unique. To produce a map we need to specify an element $f(x)$ for each $x\in\emptyset$. But there is nothing to be done! ...

Top 50 recent answers are included