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1

Well, sets and maps. In set theory, a map from $A$ to $B$ is a certain subset of $A \times B$. If the map factors through some subset $B'$ of $B$, we get two maps $A \to B$ and $A \to B'$ which are equal as sets. For example, for a set theorist, $\emptyset$ is a map from $\emptyset$ to any set. But for a category theorist, a morphism has to have a specified ...


2

The trick is to embed the category of abstract simplicial complexes inside the category of symmetric simplicial sets (= functor $\mathbf{F}^\mathrm{op} \to \mathbf{Set}$, where $\mathbf{F}$ is the category of positive finite cardinals): this can be done by sending an abstract simplicial complex $X$ to the symmetric simplicial set ...


0

Thanks to Zhen Lin to bring me on the right track. The misunderstanding is the following: The dual object ($Y$ in my example) is unique up to isomorphism, but this isomorphism is not unique. However, a dual is more than merely the object, it's the triple $(Y,\epsilon,\eta)$. This one is indeed unique up to unique isomorphism, and the calculation in the ...


1

You can define a "simplex" as having an orientation, thus getting an easier answer. A $k$-simplex is the convex hull of a set of $k+1$ points. But what does it mean for a set to have $k+1$ points? That there is a bijection from $\{1,2,\dots,k+1\}$. So simply define "$k$-simplex" in terms of a map $\{1,2,\dots,k+1\}\to\mathbb R^n$ and you can pick an ...


5

One way to formalize the idea is to say the following: the category of classical propositional theories is a reflective subcategory of the category of intuitionistic propositional theories. In order to make sense of that claim, it's best to pass from "theories" to "algebras." Each classical theory corresponds to a Boolean algebra, and each intuitionistic ...


1

Introduction to Category Theory by Harold Simmons is a nice and gentle way to get into category theory with plenty of exercises (and full solutions!). I'm an undergrad as well, and I worked through this book before moving on to Categories for the Working Mathematician because it is more leisurely. More to the point of your question, Intro to Category Theory ...


0

Thanks for asking this, I'd wanted to ask almost the exact same question, word for word! I don't have a very good answer, but I think the introductory book by Halmos and Givant 1988 ("Logic as algebra") is helpful. Is that one of the 2 books you mentioned? Knowing some algebraic geometry (such as the use of ideals) may help in understanding the book. See ...


4

Check out the new book (amazon-link) Tom Leinster, Basic Category Theory, Cambridge Studies in Advanced Mathematics, Vol. 143, 2014


5

If $S$ has a categorification $\mathrm{Decat}(\mathcal{C}) \to S$ and $T$ has a categorification $\mathrm{Decat}(\mathcal{D}) \to T$, then a categorificiation of a map $S \to T$ is a functor $\mathcal{C} \to \mathcal{D}$ such that the induced map $\mathrm{Decat}(\mathcal{C}) \to \mathrm{Decat}(\mathcal{D})$ makes the diagram $$\begin{array}{cc} ...


0

An alternative: Category theory for applied scientists: You find links and commentary here: http://johncarlosbaez.wordpress.com/2013/05/23/category-theory-for-scientists/ (A version is also posted on arXiv)


5

I can recommend Category Theory by Awodey for your situation. It is more elementary but not too slow for an average undergraduate.


4

A more elementary text is Conceptual Mathematics: A First Introduction to Categories. Also interesting is the list given here.


2

Your observation is correct, intersections are just special cases of limits. In the definition of an intersection (Def. 11.23 in Joy of Cats) it is not assumed that the indexing set $I$ is small. Therefore if a category has small limits, it doesn't need to have all intersections (only small ones). In well-powered categories this doesn't make any difference ...


2

These are "uninteresting" in the following sense. The pullback of a pushout diagram $A \leftarrow B \rightarrow C$ is the identity arrow $B \to B$ (exercise). The dual statement for pullback diagrams is true by duality. The point is that these diagrams, in these directions, aren't imposing an interesting universal property.


1

It is almost equivalent to the condition that p is an extremal epimorphism. A few months ago I made some notes about this condition: http://arxiv.org/abs/1311.2974 (related MO question: http://mathoverflow.net/questions/143070/two-pullback-diagram)


-1

Filtered colimits commute with finite limits. A filtered colimit is a colimit on a filtered diagram category, e.g. on a directed set.


3

As the author's references and abstract suggest, nearness spaces are a slight generalization of proximity spaces, an older concept. There are a handful of Questions at Math.SE about proximity spaces. Here are some "recent" research papers connected with nearness spaces: Khare, Mona and Surabhi Tiwari: "Approach Merotopological Spaces and their ...


1

It seems to me that patches or coordinate charts require some choice of basis. However, the real thing which defines a hyperplane is that it has one less dimension than the ambient vector space. However, manifold dimension is not quite restrictive enough as there are plenty of curved spaces of codimension $1$. We should also insist that it is possible to ...


2

Something that deserves to be said here: $\mathsf{Rel}$, unlike $\mathsf{Set}$, is far from complete or cocomplete! Below I'll give an example of an equalizer which fails to exist in $\mathsf{Rel}$. Since $\mathsf{Rel}$ is self-dual, it also doesn't have coequalizers. The picture does not improve much if we consider $\mathsf{Rel}$ as a 2-category: it has ...


4

The category of fields is not well-behaved. There is no free extension field on given generators. You have to specify the relations between the generators, as well as the minimal polynomials of these generators. But even if you do that, it's not always possible to find a universal solution. If $K$ is a field, then $K(x)$ may be described as a field equipped ...


2

In the category of extension fields of $R$, the morphisms $Q(x) \rightarrow S$ correspond bijectively to the elements of $S$ that are transcendental (over $Q$). So if you define $\mathcal{R}$ to be the functor that sends an extension field of $R$ to the set of elements that are transcendental (over $Q$), then its left adjoint $\mathcal{L}$ has the property ...


1

In the case of rings and polynomials, the universal property of $R[x]$ is that for any ring $R'$ with a homomorphism $\phi:R\to R'$ and any $y\in R'$ there is a unique homomorphism $R[x]\to R'$ with $x\mapsto y$. But the elements of $\mathbb Q(x)$ are rational functions of $x$, which means they can't be evaluated everywhere - there is no "evaluation at $y$" ...


0

The basic observation is that for every set $E$ we have the following comonoid structure $\Delta \colon E \to E \times E$ and $\epsilon \colon E \to *$, where $\Delta$ is the diagonal morphism and $\epsilon$ is the unique morphism from $E$ to the terminal object (this is also the unique comonoid structure that can be posed on a set). Now applying the yoneda ...


5

The best way of understanding the algebras for a monad is to find an adjunction whose induced monad is (isomorphic to) the one you started with. In this case, one may verify that the adjunction $$E^* \dashv \Pi_E : \mathbf{Set}_{/ E} \to \mathbf{Set}$$ has induced monad (isomorphic to) the "environment" monad $(-)^E$. Unfortunately, this adjunction is not ...


0

More than a year later, here's my answer to the question posed by my old self: the concept of a monad is what you're grasping for.


1

The question is way too broad. Therefore my answer is also broad. Let me only give some key notions which fit to the mentioned examples.. Creation of limits Internal hom Topological category Grothendieck fibration


3

As noted by Zhen, complements cannot be described via limits or colimits. Nevertheless, we have the following interpretation: Recall the notions of monoidal categories and internal homs in monoidal categories. In a monoidal poset, $\underline{\hom}(x,y)$ is the largest object such that $ \underline{\hom}(x,y) \cdot x \leq y$. If $y=0$ is an initial object ...


4

Yes. By the the explicit construction of the fiber product of locally ringed spaces (in particular, of schemes), it follows that the continuous map $|X \times_Z Y|\to |X| \times_{|Z|} |Y|$ is surjective and that the fiber over some point $(x,y,z)$ in $|X| \times_{|Z|} |Y|$ is $\mathrm{Spec}(\kappa(x) \otimes_{\kappa(z)} \kappa(y))$. A basis of the ...


10

Let me work only with groups for the sake of familiarity. The wreath product is defined using the semidirect product, so the first step is to understand the semidirect product categorically. I claim that the correct way to understand the semidirect product is not categorically but higher-categorically, and also that the correct higher-categorical ...


-1

I'll put up a resolution proof here. It probably doesn't help. assumption 1 ANac assumption 2 ANbc assumption 3 Aab R 1, 3 4 Abc R 2, 4 5 c


3

Your argument is not correct. In fact, $F$ is fully faithful. Define a functor $G : \mathsf{Set} \times \mathsf{Set} \to \mathsf{Set} / \{a,b\}$ by mapping two sets $A,B$ to the function $G(A,B) : A \coprod B \to \{a,b\}$ which maps the elmenets of $A$ to $a$ and the elements of $B$ to $b$. Here, $A \coprod B= A \times \{1\} \cup B \times \{2\}$ is the ...


2

In many concrete categories, the initial object is either the empty object, or some similar thing like the free object on no generators. In many concrete categories, the terminal object is the one-point set.


1

I'm afraid your argument is not correct: there are morphisms from $f$ to $h$. Hint: Perhaps $F$ is not an $isomorphism$ of categories, but something slightly weaker?


2

Given a natural transformation $$\varphi_{} \colon \mathbf A(-,G(-)) \to \mathbf B(F(-),-) $$ this is the same as a family natural transformation between the functors $$\varphi_Y \colon \mathbf A(-,G(Y)) \to \mathbf B(F(-),Y)$$ natural in $Y$. For every $Y \in \mathbf B$ by yoneda lemma we have that, if $\epsilon_Y=\varphi_Y(1_{G(Y)})$, then ...


0

Actually, the definition of adjunctions via natural isomorphisms of Hom-functors works in any $2$-category. Specifically, $f\leftrightarrows g$ are adjoint with unit $1\stackrel\eta\Rightarrow fg$ and counit $gf\stackrel\epsilon\Rightarrow 1$ if and only if we have a family of isomorphisms $[fX,Y]\cong[X,gY]$ that is natural in the "$E$-objects" $E\stackrel ...


2

Here is a way to come up with the exponential objects. Assuming that $C$ is nice enough, the forgetful functors $I,O,S : \mathrm{Aut}(C) \to C$ should have left adjoints $I^*,O^*,S^*$. For example, for every object $X \in C$, then $I^*(X)$ will be the free automaton generated by an input object $X$. Explicitly, this is given by input object $X$, state ...


1

$\Sigma^{-1} A$ represents the functor of homomorphisms on $A$ which invert all elements in $\Sigma$. Since $\Sigma$ is covered by the $S_i$, this means that all the $S_i$ are made invertible. Hence, we get unique lifts to $S_i^{-1} A$, and these are clearly compatible. The construction is invertible, so that $\Sigma^{-1} A$ represents the same functor as ...


1

Let $A=\left(\begin{array}{cc}1&0\\0&2\end{array}\right)$, $B=\left(\begin{array}{cc}0&1\\2&0\end{array}\right)$, $C=\left(\begin{array}{cc}2&0\\0&1\end{array}\right)$, $D=\left(\begin{array}{cc}0&2\\1&0\end{array}\right)$. Then $ABCD=\left(\begin{array}{cc}1&0\\0&16\end{array}\right)$ and ...


1

Recall $K(\alpha)=K[x]/(f)$ and that polynomial algebras have a universal property (freely adjoin an element) and quotient rings have a universal property (kill elements in the ideal). It follows that for any $K$-algebra $R$ we have a bijection between $K$-algebra homomorphisms $K(\alpha) \to R$ and roots of $f$ in $R$. And yes, this is a universal property, ...


4

"Uniqueness will fail for noncommutative monoids, so here we must use commutativity of our monoids, " This is not correct. You get a morphism in $V$ also for non-commutative monoids, but it won't be a monoid morphism. This is where commutativity is used. Uniqueness holds in general: If $h : M \otimes N \to X$ is a monoid morphism with $h \circ (M \otimes ...


3

The universal property is that $K(\alpha)$ is the universal (initial) extension of $K$ together with a root of the minimal polynomial of $\alpha$. The algebraic closure has the property that every polynomial has a root, so it is in particular such an extension. (One might think it's the algebraic closure that's supposed to have a universal property here, but ...


2

Note that the result is true in any category (not necessarily additive) having finite (co)products, (co)equalisers for any pair of parallel of arrows and a terminal (initial) object. More generally again, a category admitting arbitrary small (co)products and (co)equalisers for any pair of parallel arrows is (co)complete. This is a completely standard fact ...


2

At least two different terms are used in the literature for a commutative monoid in which division is a partial order: holoid and naturally partially ordered. Another possibility would be $\mathcal{H}$-trivial since a commutative semigroup has the required property if and only if the Green's relation $\mathcal{H}$ is the equality in this monoid. See ...


1

Well, to show that $f_!$ takes acyclic complexes of soft sheaves to acyclic complexes, we need to split up an acyclic complex of soft sheaves $$0 \to \mathcal F_1 \xrightarrow{\phi_1} \mathcal F_2 \xrightarrow{\phi_2} \mathcal F_3 \xrightarrow{\phi_3} \mathcal F_4 \xrightarrow{\phi_4} \dots$$ into short exact sequences, namely $$0 \to \mathcal F_1 ...


8

Let $0_M$ and ${\rm id}_M$ denote the zero map and identity map on an $A$-module $M$. We have $$M=0\iff 0_M={\rm id}_M.$$ Since $F$ is a functor, $F({\rm id}_M)={\rm id}_{FM}$. Since it's also additive, $F(0_M)=0_{FM}$.


3

Question 0: Here's a reasonable definition. For every finite subset $K'\subseteq K$, we can view $F(K')$ as a subalgebra of $F(K)$ (the inclusion map is obtained by applying the free functor to the inclusion map of sets $K'\hookrightarrow K$). For $t\in F(K)$, $\text{supp}(t)$ is the smallest subset $K'$ such that $t$ is in $F(K')$. I'll leave it to you to ...


3

For subsets of the natural numbers, and for sets naturally connected with such subsets via indexing, the subject has been studied for a very long time. The Wikipedia article on the Arithmetical Hierarchy will give a start. There has been particularly fine-grained analysis of r.e. degrees, huge literature.


-1

You may wish to extend your ordering range to the ordinals, which are in some senses more fine-grained than cardinals. You may also wish to read about order-embeddings and the constellation of related ideas that are linked and referenced there.


0

The following is what I have after running through "A Course in Universal Algebra" by Burris et al. using the description of a group given here. It's just a sketch. Consult the text wherever necessary. We work in the type $\mathcal{G}=\{\cdot, \sim, e\}$, where $\cdot\in\mathcal{G}_2$ is a binary operation, $\sim\in\mathcal{G}_1$ is a unary operation, and ...


0

You can prove it in Sequent Calculus as described in the pdf mentioned in the comments (http://zll22.user.srcf.net/talks/2011-12-01-CategoricalLogic.pdf ) 1 A |- A Identity 2 B |- B Identity 3 C |- C ...



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