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2

If somebody referred to a full subcategory of an abelian category closed under extensions, then I'd assume that they probably meant the subcategory to be additive, and therefore containing a zero object. But you could, for example, take $\mathcal{A}$ to be the category of finite dimensional vector spaces over a field, and $\mathcal{B}$ to be the full ...


2

A mapping $f:X\to X$, for a set $X$, which satisfy $f\circ f=1$ is called involution.


4

It simply doesn't follow that $g$ is a morphism in $S$. For example, $A$ might be the category of abelian groups, $S$ might be the full subcategory of finitely generated abelian groups, and the target of $g$ might be an infinitely generated abelian group. But the proof is very easy to repair: just define $g = \text{coker}(f) \in A$ in the first place. Then ...


4

I contend that we can reconstruct $\mathbb Z_{>0}$ up to a permutation of the form $$\prod_i p_i^{a_i}\mapsto \prod_i \pi(p_i)^{a_i}.$$ Since such maps are homomorphisms of the multiplicative monoid we can reconstruct multiplication. We can associate with every $x$ the set of proper divisors, i.e., $D(x):=\{\,y\in\mathbb Z_{>0}:y\ne x, y\mid x\,\}$. ...


15

No. The reason is that the splitting field is not unique up to unique isomorphism, and any terminal object has to be unique in this stronger sense.


0

Probably no longer useful, but I find that Murre's Introduction to Grothendieck's theory of the etale fundamental group is quite an excellent source. http://www.math.tifr.res.in/~publ/ln/tifr40.pdf


5

The categories are not even equivalent. In $\mathbf{Set}$ there exists an arrow from a terminal object to a non-terminal object -- for example one of the functions $\{1\}\to\{1,2\}$. But $\mathbf{Set}^{op}$ does not have this property, because such an arrow in $\mathbf{Set}^{op}$ would be a function from a nonempty set to the empty set.


3

Yes. In SET singletons are terminal objects. Then in its opposite category any singleton is an initial object. However in SET there is only one initial object (the empty set). This shows that the two categories are not isomorphic.


1

Let $C$ be any category whatsoever, and let $c \in C$ be an object. I claim that the representable functor $\text{Hom}(c, -) : C \to \text{Set}$ has a left adjoint iff all coproducts $\bigsqcup_S c$ exist, and that when this is true the left adjoint sends a set $S$ to $\bigsqcup_S c$. Theorem 3.4.5 guarantees that the $C$ you care about is cocomplete, so in ...


1

Here is a possible starting point. $\text{Grp}$, as well as other familiar categories of algebraic objects like $\text{Vect}$ or $\text{Ring}$, are distinguished from arbitrary categories by the fact that they are categories of models of Lawvere theories in $\text{Set}$. This is a categorical way of talking about universal algebra. A categorical ...


2

The question reduces to a question about sheaves of sets, so I will just work with those (instead of modules or whatever). The point is that we have the following commutative diagram of right adjoint functors, $$\require{AMScd} \begin{CD} \mathbf{Sh}(X) @>>> \mathbf{Sh}(Y) \\ @VVV @VVV \\ \mathbf{Psh}(X) @>>> \mathbf{Psh}(Y) \end{CD}$$ ...


2

Normally this kind of thing is verified by the "what else?" argument, that is, what are the chances you really have two different canonical maps in that square? That said, it could be comforting to know it's possible to really check such a thing. So: Let me change your $\oplus_i$ to $\oplus_j$, to avoid collision of notation. We have to show that for every ...


1

First example. Let $\mathrm{Mor}(x,y)$ be empty for $x<y$, and consists of one element if $x\ge y$. So, if you can write such a diagram, $(a_n)$ is non-decreasing sequence. Therefore this diagram has the inverse limit iff $(a_n)$ bounded above, and inverse limit equals to supremum. Second example. Let $\mathrm{Mor}(x,y)$ be the set of all paths ...


29

Well. That depends on whom you might ask this. Set theory might be inconsistent. In particular $\sf ZFC$ and its extension by large cardinal axioms. It's a nontrivial thing, to feel safe with these theories, and it takes a lot of practice and time until you understand that $\sf ZFC$ is self-evident to some extent, and [some] large cardinal axioms are ...


7

If you're looking for motivation to pursue other foundations, I recommend the article "Rethinking Set Theory" by Tom Leinster: http://arxiv.org/pdf/1212.6543v1.pdf. In particular, he's providing a gentle, well-motivated introduction to Lawvere's Elementary Theory of the Category of Sets (ETCS). Leinster mentions at least two complaints. First, in ...


-5

Yes. Category theory is one example of a theory than transcends ZFC. Some other examples include Hyperreals and Surreals. There are many more.


1

We have answer using projections to prove universal property by Peter Smith, and I'm going to show another way to prove this, although, depending how one defines products (limits), this may actually be longer to work out from scratch. But, the characterization of product I'm going to use is very useful in itself: $P$ is product of the family ...


0

I've finally realized it works the same way for any $g$ as it works for $f$ That is, we get from the universal property of $(A \times B) \times C$ that $\forall$ morphisms $f_1: X \to A \times B$ there is a unique morphism $\phi: X \to (A \times B) \times C$ such that $\pi'_{A \times B}\phi = f_1$ But it follows that for all morphisms $\pi_Af_1 = f, ...


2

Theorem: $(X_1 \times X_2) \times X_3$ together with the obvious projection arrows forms a ternary product of $X_1, X_2, X_3$. Proof Assume $[X_1 \times X_2, \pi_1, \pi_2]$ is a product of $X_1$ with $X_2$, and $[(X_1 \times X_2) \times X_3, \rho_1, \rho_2]$ is a product of $X_1 \times X_2$ with $X_3$. Take any object $S$ and arrows $f_i\colon S \to ...


2

You don't need to use the fact that $φ$ is monomorphism any more - you already did that by assuming $Y → Y ×_Z Y$ is isomorphism, which is equivalent to $φ$ being mono. All that is left to do is to observe that the bottom square is also cartesian (for trivial reasons), which makes the whole outer square cartesian too. Now $g$, being a pullback of an ...


5

I'm afraid, that this question may have a really easy answer, however... Unfortunately it does. Assume $A×B ≅ 1$ (the terminal object). Then for every $C$ we have $$1 ≅ \mathrm{Hom}(C, 1) ≅ \mathrm{Hom}(C, A × B) ≅ \mathrm{Hom}(C, A) × \mathrm{Hom}(C, B),$$ so $\mathrm{Hom}(C, A)$ and $\mathrm{Hom}(C, B)$ are both one-element sets, and $A$ and $B$ ...


1

The $A$-algebra $S^{-1}A$ is a member of this category: an element of the multiplicative subset $s \in S$ is a unit as it is inverted by $1/s$. It is furthermore initial among these algebras because the unique morphism$$\overline{\varphi}: S^{-1}A \to B$$is given by$$\overline{\varphi}\left({r\over{s}}\right) = \varphi(r)\varphi(s)^{-1},$$where $\varphi: A ...


0

The answer, if anybody is still wondering, is simplicial homotopy.


1

You can use associativity, and the the bialgebra condition to get $$ (\nabla\otimes \operatorname{id}\otimes \operatorname{id})(\operatorname{id}\otimes \Psi\otimes \operatorname{id})(\Delta\otimes \Delta\nabla), $$ and then your identity to eliminate the braiding only once.


2

Another example, concrete in a different, informal sense than Hagen von Eitzen's example: Let $\mathcal B$ be the partially ordered set, viewed as a category in the usual way, consisting of three elements, say 0,1,and 2, with $0\leq1$ and $0\leq 2$ while 1 and 2 are incomparable. Let $\mathcal C$ be the partially ordered set obtained by adjoining to ...


1

As a concrete example, you may know that the coproduct of groups is the free product and the coproduct of abelian groups is just the direct sum. The free product is never abelian (unless one factor is trivial), hence the inclusion from Ab to Grp certainly fails to preserve coproducts.


3

In fact, every partially ordered set can be viewed as a category in the way you describe; your example is obtained by applying this construction to the poset $(\mathbb{Z}_{\geq 1},\mid)$. This process always turns meets into products and joins into coproducts. So your observations are essentially the fact that $\mathrm{gcd}$ is the meet in $(\mathbb{Z}_{\geq ...


4

The problem here is that $\mathcal C$ does not know what "multiplication" is. The entity $ab$ is entirely meaningless to it. It is similarly meaningless to say that $\dfrac{ab}\ell$ is the GCD, because as far as $\mathcal C$ is concerned, this object is not connected to $a$, $b$ or $\ell$. The only thing that $\mathcal C$ can hope to express things about ...


1

I'll use the terminology of Hovey's book on Model Categories. If $X$ is an $R$-module, and $\iota:X\to I$ a monomorphism from $X$ to an injective module, then $X\oplus I$ is a cylinder object for $X$ with the maps $$X\oplus X\stackrel{\begin{pmatrix}1&1\\\iota&0\end{pmatrix}}{\to}X\oplus I\stackrel{\begin{pmatrix}1&0\end{pmatrix}}{\to}X.$$ All ...


1

Let's break it down. A natural isomorphism is a natural transformation that is also an isomorphism. Or alternatively, a isomorphism (technically a family of isomorphisms) that is natural. Associative "up-to natural isomorphism" means, somewhat informally, that $(A\otimes B)\otimes C$ is not necessarily equal to $A\otimes (B\otimes C)$, but there is an ...


1

Yes. They both refer to the hyperconnected–localic factorisation of geometric morphisms.


2

$\newcommand{\CC}{\mathscr{C}}\DeclareMathOperator{\ob}{ob}\DeclareMathOperator{\mor}{mor}\newcommand{\RR}{\mathscr{R}}$The "2-category" aspect is a bit of a red herring in my opinion. It may be easier to first consider the case where $\CC$ is a category, and $\RR(X)$ is a set for all $X$. But I'll do the case of 2-categories anyway. Let ...


2

We're assuming that both $A$ and $B$ are symmetric $R$-algebras, which means that $A\cong\operatorname{Hom}_R(A,R)$ as $A$-bimodules, and similarly for $B$. And we're assuming that $_AM_B$ is an $(A,B)$-bimodule finitely generated and projective over $A$ and over $B$. For left $A$-modules $_AX$ and $_AY$, there is a natural $R$-module homomorphism ...


1

This sheafification does not exist in general, at least if you want the target category to be the category of compact spaces. To construct the sheafification of a sheaf, you need the existence of certain limits and colimits, and the category of compact spaces does not have all limits and colimits (in particular, it does not have all equalizers and does not ...


1

Suppose $X$ is the open ball in $\mathbb{R}^n$. Denote by $\mathcal{C}$ (for compactification) the sheaf you defined, and $\mathcal{C}^s$ its sheafification. For a point $x \in X$ and balls $U_n$ centered at $x$ of radius $1/n$ we have the stalk of $C^s$ at $x$ is given by $\mathcal{C}^s_x = colim_{n \rightarrow \infty} \mathcal{C}(U_n)$. The colimit on ...


2

This is only a partial answer, but it's too long for a comment, so I'll post this as an answer and hope that nobody complains. For simplicity, let's call your presheaf $\mathcal{F}$. First, let's look at the stalk: The stalk $\mathcal{F}_x$ is just going to be the direct limit (with respect to the restriction you defined) of the spaces $U^*$ with $U$ an ...


6

Denote $F$ for the free category functor and $U$ for the underlying graph functor. Let $\mathcal C$ be a small category. Then the morphisms of $FU(\mathcal C)$ are finite sequences of composable morphisms in $\mathcal C$. One can put an equivalence relation $\sim$ on the morphisms of $FU(\mathcal C)$: it identifies $(f_1,\dots,f_n)$ and $(g_1,\dots,g_k)$ ...


3

The equivalence between the category of compact Riemann surfaces with nonconstant holomorphic maps and the category of function fields over the complex numbers of transcendence degree one with morphisms of complex algebras can be found in Chapter 1.3 of 'Introduction to Compact Riemann Surfaces and Dessins d’Enfants' by Ernesto Girondo and Gabino ...


1

This is perhaps not the most natural example, but the category of groups doesn't have a cogenerating set (se here for why), so $\mathrm{Grp}^{\mathrm{op}}$ can't have a generating set.


5

It depends on what you mean by "generators." There is a notion of a collection $S$ of objects being a "family of generators" of a category $C$, which means that the functors $\text{Hom}(s, -) : C \to \text{Set}, s \in S$ are jointly faithful. Any category has a (possibly large) family of generators given by taking every object, so the interesting question is ...


1

This is not true in general. Instead, the equation $$ \Psi^{-1}=(\nabla\Psi^{-1}\otimes \nabla\Psi^{-1})(S^{-1}\otimes \Delta\nabla\Psi^{-1}\otimes S^{-1})(\Delta \otimes \Delta) $$ holds. This may not be very useful as $\Psi^{-1}$ appears on either side. Note that $S^{-1}$ only satisfies the twisted convolution invertibility $$ ...


1

Since others have given detailed answers, at appropriate math.se length, I won't try to do that. But as Clive says, the details can seem reasonably grim at first sight. So I'll just add that if you want a more slow-motion version, I have a bash at making the full-dress Yoneda Lemma, including the naturality claims, as clear as I can in these draft Notes on ...


1

We can actually exhibit the isomorphism $\phi $ in $\tag1Nat(h^A, F) \cong F(A)$. For any natural transformation $\nu :h^{A}\to F$, define $\phi :Nat(h^A, F)\to F(A)$ by $\phi (\nu)=\nu _{A}(1_{A})$. Note that $\nu _{A}$ is simply the $A$ component of $\nu $. i.e. $\nu _{A}$ is a morphism, in fact a $set$ map: $\hom(A,A)\to FA$ so that $\phi $ sends ...


4

The details are reasonably grim but if you stare at them hard enough you'll see that the morphisms are the only things they could possibly be. The actions on objects of the two functors $\mathsf{Set}^{\mathcal{C}} \times \mathcal{C} \to \mathsf{Set}$, respectively, are $$(F,A) \mapsto \mathsf{Nat}(y^A, F) \quad \text{and} \quad (F,A) \mapsto F(A)$$ The ...


2

A morphism $\left(F, A\right) \to \left(G, B\right)$ in $\operatorname{Set}^C \times C$ has the form $\left(\alpha, p\right)$, where $\alpha : F \to G$ is a morphism in $\operatorname{Set}^C$ (that is, a natural transformation $F \Rightarrow G$) and where $p : A \to B$ is a morphism in $C$. For every morphism $p : A \to B$ in $C$, there is a natural ...


0

The category you look for is the cograph of the profunctor $U^*:{\bf Set}^{op}\times{\bf Grp}\to{\bf Set}$ induced by the forgetful functor $U:{\bf Grp}\to{\bf Set}$, i.e. $U^*(S,G)=\hom_{\bf Set}(S,UG)$. It basically connects the categories ${\bf Set}$ and ${\bf Grp}$ with further morphisms from sets to groups: an arrow $S\to G$ will be just a function ...


1

Fortunately this has little do to with bifunctors. First of all, note that limits are functorial: given two diagrams $D$, $D' : \mathscr I → \mathscr D$ and a natural transformation $α : D → D'$, we can get a morphism $\lim α : \lim D → \lim D'$. This is easy to construct and write out in some explicit examples of limits (eg. try to do it for products). In ...


5

It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity. So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;S^{-1}\varphi\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there ...


4

What you are looking for is a particular instance of the notion of localization of a category with respect to a set of morphisms: Given a category ${\mathscr C}$ and a set of morphisms $S\subset\textsf{Mor}({\mathscr C})$, you ask for the inital functor ${\mathscr C}\to{\mathscr D}$ mapping the morphisms in $S$ to isomorphisms. Ignoring set theoretical ...


6

Your idea is basically correct. More explicitly, $F\mathcal{A}$ has the same objects as $\mathcal{A}$ and a map $A\to B$ in $F\mathcal{A}$ can be represented as a "zigzag" of maps of $\mathcal{A}$ $$A\to C_1\leftarrow C_2\to C_3\leftarrow\dots\leftarrow C_n \to B,$$ which we think of as formally representing the composition of the rightward arrows with the ...



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