New answers tagged

0

Note that since the kernel pair $K[f]\rightrightarrows X$ of a morphism $X\xrightarrow{f}Y$ is really the cartesian square of $X\xrightarrow{f}Y$ in the slice category $\mathcal C/Y$, we have a functor $K\colon\mathcal C/Y_K\to\mathcal C/Y$ where $\mathcal C/Y_K$ is the full subcategory of $\mathcal C/Y$ generated by morphisms that have kernel pairs. ...


-1

Let consider a category with equalizers. If a morphism $f:A\to B$ has image $m:M\to B$, then $m$ is a monomorphism and there exists a unique morphism $e:A\to M$ such that $f=m\circ e$ and $e$ is an extremal epimorphism (that's it doesn't factor trought any monomorphism). For let $a,b$ be a parallel pair of morphisms such that $a\circ e=b\circ e$ and let $q$ ...


2

Like Geoff suggests in the comments, you can use $\text{Ker}$ for the object and $\text{ker}$ for the morphism. But I don't think this convention is universal so you should probably say that you're using it. I sometimes use $\text{ker}$ for both.


0

This is not true. Let $A=C[0,1]$. Consider the increasing sequence of closed sets closed sets $F_0=\varnothing$, $F_i=[0,2^{-i}]$, and set $I_i=\left\{f\in A:f|_{F_i}=0\right\}$. Then we are in the setting you described. We can identify $A/I_i$ with $C(F_i)$, via the map $A/I_i\to C(F_i)$, $f+I_i\mapsto f|_{F_i}$. With this $A/I_{i+1}\to A/I_i$ is ...


0

I recommend Goldbatt's Topoi for a first introduction. Section 16. tells about geometric morphisms.


0

It is conventional to say that a category is $\kappa$-small if it has $<\kappa$ (objects and) morphisms. So $\aleph_0$-small means finite, since anything strictly smaller than $|\Bbb N|$ is finite. And therefore an $\aleph_0$-limit, since it is a limit over a diagram whose domain is an $\aleph_0$-small category, recovers precisely the notion of finite ...


3

The term "universal property" does not refer just to limits. More generally, it refers to any structure that can be described as a terminal object in some category (probably not your original category). For instance, the limit of a diagram is just a terminal object in the category of cones under the diagram. In this case, the category in question is the ...


3

Here the adequate notion is that of adjoint functors : there is the inclusion functor $U: Field\to Domain$ (the functor that forgets that a field is more than an integral domain). The functor $Q: Domain\to Field$ that takes an integral ring to its field of fraction is the left adjoint to $U$ : you have $Hom(Q(R),F) = Hom(R,U(F))$ for any domain $R$ and any ...


1

I can give you a Haskell answer, though this is this maybe somewhat off-topic on a math board. In Haskell, the trivial structure aside, there are things that are called as "Readers" and "Writers" that are both Monads and Comonads. Intuitively, while the Monadic structure allows us to "smush"layers of functors together (since the Monad is a monoid of ...


1

It perfectly makes sense to talk about the inverse image functor in the full category of sets. Let $f \colon X \to Y$ be a set-theoretic map: by the universal property of pullbacks, it induces an ajunction $$ f_\ast : \mathsf{Set}/X \rightleftarrows \mathsf{Set}/Y : f^\ast $$ where the left adjoint $f_\ast$ is just the postcomposition by $f$ andthe right ...


2

Calling this category the homotopy category at this level of generality is a bit misleading. The sense in which it is a homotopy-category-as-in-morphisms-up-to-homotopy comes from, for example, taking the simplicial localization first. This is a simplicially enriched category which presents an $\infty$-category. It has a notion of homotopy between maps given ...


4

No, a coproduct is not a filtered colimit in general, because your set $S$ is interpreted as a discrete category, and a discrete category is not filtered if it has at least two objects : take $x,y$ two different objects. If the category was filtered you would have some $z$ with morphism $x\to z$ and $y\to z$, but that's impossible because that implies $z=x$ ...


2

A weighted (co)limit in an arbitrary (co)complete enriched category can be expressed as a (co)end. The formulas are:$$ F\star G \cong \int^C FC\otimes GC \qquad \{F,G\} \cong \int_C [FC,GC] $$ where I'm using $F\star G$ as the weighted colimit, $X\otimes C$ as the tensor, and $[X,C]$ as the cotensor. If you look in Basic Concepts of Enriched Category ...


1

You should read again: it says that it is a functor $\varphi\colon \mathcal A \to \mathcal B^{\to}$ where $\mathcal B^\to$ is the arrow category of $\mathcal B$. The function $\operatorname{Ob}\mathcal A \to \operatorname{Mor}{\mathcal B}$ you are talking about is only the object-mapping part of the functor. The naturality condition is actually taken into ...


2

A contravariant adjunction on the right consists of contravariant functors $F : \mathcal{C} \to \mathcal{D}$ and $G : \mathcal{D} \to \mathcal{C}$ and a natural bijection $$\mathcal{D} (Y, F X) \cong \mathcal{C} (X, G Y)$$ where $X$ varies in $\mathcal{C}$ and $Y$ varies in $\mathcal{D}$. Note that we can equally well think of this as an ordinary (or ...


2

Let $\mathsf{C}$ be a linear category. To give $\mathsf{C}$ the structure of a $\mathsf{Vect}$ (left) module essentially amounts to giving a bifunctor $$\otimes : \mathsf{Vect} \times \mathsf{C} \to \mathsf{C}$$ plus associativity and unitality constraints satisfying a bunch of conditions. I will only give an idea of the definition of $\otimes$, constructing ...


8

There is no group object: since you only have bijections in your category, for all G there exists an X such that hom(x,G) is empty, so not a group in any way at all.


-1

The question uses a nonstandard definition of group object. Under the ordinary definition of group object, the answer to the first question is For the identity morphism $1 \to G$ and the multiplication morphism $G \times G \to G$ (if products exist...) to live in a category whose morphisms are bijections, $G$ must have cardinality $1$. The definition in ...


2

The triangle identities imply that $$G\epsilon F \circ GF\eta=id_{GF} =G\epsilon F\circ \eta GF$$ and $$FG\epsilon \circ F\eta G= id_{FG}=\epsilon FG\circ F\eta G.$$ Thus $FG\epsilon=\epsilon FG$ and $GF\eta =\eta GF$ as soon as $\epsilon$ or $\eta$ is an isomorphism. Since $\epsilon$ (resp. $\eta$) is an isomorphism if and only if $G$ (resp. $F$) is fully ...


2

The $\text{Hom}$ functor preserves limits in each argument (in a very strong sense), neither preserves colimits in general. You should prove this as it's the source of continuity for most other things that are continuous, most notably adjoints. (Because (co)limits in functor categories are computed point-wise, this lifts to the Yoneda embeddings.) But be ...


6

Let $[n]$ denote the category $0\to 1\to\cdots\to n$. Given a factorization system $(L, R)$ in a category $\mathcal{C}$, we can choose a map $F_0\colon\operatorname{Ob}\mathcal{C}^{[1]}\to\operatorname{Ob}\mathcal{C}^{[2]}$ which maps an object $x\xrightarrow{f} y$ in $\operatorname{Ob}\mathcal{C}^{[1]}$ to an object $x\xrightarrow{e} E\xrightarrow{m} y$ in ...


1

Here is an example of an equivalence of categories $F : \mathbf{C} \leftrightarrows \mathbf{D} : G$ where the natural isomorphisms $1_{\mathbf{C}} \overset{\eta}{\to} GF$ and $FG \overset{\epsilon}{\to} 1_{\mathbf{D}}$ do not give an adjunction. Let $G = \mathbb{Z}/3\mathbb{Z}$, and let $\mathbf{C} = \mathbf{D} = \mathbf{B}G$ the category with one object ...


0

Left adjoints preserve colimits; right adjoints preserve limits. To ask that a functor preserves both limits and colimits is a fairly strong condition; in the context of abelian categories this is even stronger than being exact, which means preserving just finite limits and colimits. And examples of nonexact left adjoints between abelian categories abound: ...


0

Given a trilinear map $\mu:M_1\times M_2\times M_3\to T$, for each $m_3\in M_3$ the map $(m_1,m_2)\mapsto \mu(m_1,m_2,m_3)$ is an $(A_0,A_1,k)$-bilinear map $M_1\times M_2\to T$. There is thus a unique map $\nu:(M_1\otimes_{A_1} M_2)\times M_3\to T$ such that $\nu(m_1\otimes m_2,m_3)=\mu(m_1,m_2,m_3)$ and $\nu$ is $A_0$-linear in the first variable. It is ...


1

I'll suppress superscripts and subscripts, write $[-,-]$ for $\mathbf{Hom}(-,-)$, and write $i^*$ as $[i,Y]$, etc. The first diagram involves the inclusion $r$ of the horn and a map $a:\Lambda\to [L,X]$, maps $b,c:\Delta\to [K,X],[L,Y]$, and $([i,X],[L,p])$. We have $([i,X],[L,p])a=(b,c)r$, so by the universal property of the pullback $$(1)\quad ...


0

In my opinion, it becomes clearer if you take the homotopical point of view. Denote by $\mathcal I$ the interval category: its objects are 0 and 1, and it has a unique non-identity morphism $0\to 1$. Then a natural transformation between $F,G \colon \mathcal C \to \mathcal D$ is just a functor $\alpha \colon \mathcal I \times \mathcal C \to \mathcal D$ ...


1

If I understand correctly, you want to take advantage of the curryfication of the maps $G\times X \rightrightarrows X$ to define a groupoid $G \rightrightarrows \operatorname{Bij(X)}$. This can not be done. I will try to explain why. Recall that the curryfication of an arrow $f\colon A\times B \to C$ is given by $$ \bar f \colon A \to C^B \colon a \mapsto ...


0

This might not be the answer you are looking for, but it could give you a little insight on what monadic really means. Given a monad $(T,\eta,\mu)$ on a category $\mathcal C$, you can form the category $\operatorname{Adj}(\mathcal C,T)$ of adjunctions above $T$: its objects are the ajdunction $F: \mathcal C \rightleftarrows \mathcal D :U$ ($F$ on the ...


1

One does not need to invoke the axiom of choice to show that the category of groups is not small. One can merely observe that there is no largest cardinality of a group. (In particular, for any cardinal number, there is a group of larger cardinality.) It is irrelevant whether there is a group of every cardinality.


3

For an exact functor $F$ between abelian categories, being faithful is equivalent to the condition $$FX=0\Rightarrow X=0$$ on objects, since a morphism is zero if and only if its image is zero, and exact functors preserve images. So, for your second question, the "criterion" proves this equivalent formulation of faithfulness.


0

I think this answers the first part of my question (before the edit): Suppose that the map $ev:X\otimes X^\vee \to \mathbf{1}$ is not an epimorphism. Then there exists an object $Z$ and distinct morphisms $\alpha, \beta: \mathbf{1}\to Z$ such that $\alpha\circ ev = \beta\circ ev$. Thus $ev: X\otimes X^\vee \to \mathbf{1}$ factors through the monomorphism ...


4

Here the answers to your questions: Clearly not. If you take a generic sub-category of $\mathcal D$ there is no reason why it should be complete: for instance you could take the full sub-category of $\mathcal D=Mod-\mathbb Z$ spanned by the free $\mathbb Z$-modules (that arbitrary direct sums of copies of $\mathbb Z$). This category is clearly not small, ...


3

There are obvious non-small and non-complete subcategories of $\mathcal D$, e.g. $\mathcal{D}$ itself and the full subcategory spanned by two object $a,b$ and morphisms between them. The solution set condition is also not automatically satisfied. If $\mathcal C$ is a large discrete subcategory, then no map $F(C)\to D$ can factor nontrivially through any ...


2

$F\eta$ isn't a composition of a functor and a natural transformation, it's just a natural transformation. If you have a transformation $\alpha:F\to G$, then a little examination will show that there's an induced natural transformation $HF\to HG$ for any $H$ that composes appropriately, often denoted $H\alpha$; and similarly one $FH'\to GH'$ for appropriate ...


4

Given functors $$\mathcal{A} \xrightarrow{F} \mathcal{B} \overset{G}{\underset{H}{\rightrightarrows}} \mathcal{C} \xrightarrow{K} \mathcal{D}$$ and a natural transformation $\alpha : G \to H$, we can define $\alpha_F$ (also written $\alpha F$) and $K\alpha$ as follows: $\alpha_F : G \circ F \to H \circ F$ is the natural transformation whose components are ...


1

$F\eta$ is just short for $(F(\eta_C))_{C\in Ob \mathcal{C}}$, wich defines (in this case) a natural transformation $F\Rightarrow FGF$.


3

I don't know what you mean by "colimits wrt $\times$." You need to show three things: that coproducts commute with colimits, that product distributes over colimits, and that $X \mapsto X^n$ preserves $\omega$-colimits. You haven't addressed the third point, and it's also not true that $X \mapsto X^n$ commutes with arbitrary colimits, so this is in ...


3

A category $\mathcal{C}$ in which every hom-set has exactly one element is necessarily a groupoid since the unique morphism $p:A\to B$ and the unique morphism $q:B\to A$ must compose to the morphism $\mathrm{id}_A:A\to A$ and $\mathrm{id}_B:B\to B$. Moreover, $\mathcal{C}$ is a connected groupoid since every two objects of the category have a morphism ...


4

Yes; A category in which there is at most one arrow between any two objects is just a poset. A category in which there is precisely one arrow between any two objects is just a boolean. I work up to equivalence, of course. Edit. Here's a simple way of getting a boolean $\mathrm{isInhab}(X)$ from a set $X$, which expresses the proposition "$X$ is ...


1

A set $X$ determines very special functors from finite sets to sets, contravariantly by sending $S$ to the set of functions $f:S\to X$ and, dually, covariantly. These are restrictions of representable functors on the full category of sets, so for instance the contravariantly one send colimits to limits. There's no reason at all why a general functor should ...


1

The answer to your particular question is that there's no such adjunction in general. Subcategories for which the inclusion has a left (right) adjoint are called (co)reflective, and are very common but far from universal. Vaguely, you need an optimal way to round objects of the bigger category down to objects of the smaller category, and there's no general ...


0

Palace Chan, not sure if you are still perplexed by Yoneda's lemma but if so I may be able to help. Looking at an early post you write: "I now have something like F[(Ff)u]should equal ΦY(g∘f)?". You appear to be treating F here like a function acting on the set F(X). The functorial notion can be confusing that way, but F does not act as a set function, ...


1

Before answering the question we recall that a function from $A$ to $B$ is a relation $f\subset A\times B$ which has $A$ as its domain and $(x,y),(x,z)\in f$ implies $y=z$. It allows us then to write $f:A\to B$ and $y=f(x)$. Thus, a function has three intrinsic elements: a domain, a counter-domain and an association rule. Now, given $S\subset B$, $$ ...


3

Map a function $f:X\to Z^Y$ to the function $(y,x)\mapsto f(x)(y)$ where $f(x)\in Z^Y$, i.e. $f(x)$ is a function $Y\to Z$. Map a function $f:X\to Z\times Y$ to the pair of functions $(\pi_Z\circ f,\,\pi_Y\circ f)$ where $\pi_Z,\pi_Y$ are the projections from $Z\times Y$. Can you define their inverses in a similar manner?


8

A classical example : the forgetful functor from topological spaces to sets. The left adjoint is the "discrete space" functor (sending a set $X$ to the discrete space with underlying space $X$), and the composition just gives the identity on Sets, so clearly Top is not the Eilenberg-Moore category of the monad. You can see that the forgetful functor does ...


1

If $\mathcal{C}$ is a monoidal Krull-Schmidt category with finitely many indecomposables, then this translates to $r(\mathcal{C})$ being a free $\mathbb{Z}$-module of finite rank, so clearly an interesting example will be of this form. Take the ring to be $\mathbb{Z}[i]$ and note that if this arises as $r(\mathcal{C})$ for some category as above then this ...


4

It is the definition of an invertible function. A function $f:X\to Y$ is said to be invertible if there exists a function $g:Y\to X$ such that $$\forall y \in Y, f(g(y)) = y \text{ and } \forall x\in X, g(f(x)) = x$$ If $f$ has this property, we call $g$ the inverse of $f$ and note it $f^{-1}:= g$


2

For a symmetric (or even just braided) monoidal category, yes. Proof: $$\begin{align} \mathcal{C}(A,\text{Hom}(B,C)) & \cong \mathcal{C}(A\otimes B,C) \\ & \cong \mathcal{C}(B \otimes A, C)\text{ via braiding } \\ & \cong \mathcal{C}(B, \text{Hom}(A,C)) \end{align}$$ So, $\text{Hom}(-,C)$ is adjoint to itself on the right, just as for cartesian ...


3

Here's a very concrete argument. A limit of the cosimplicial object is an $X$ which comes a priori with maps to each of the cosimplicial levels, but since $0$ is weakly initial in $\Delta$, only needs to be given a map $f:X\to \prod FU_i$, from which the other maps are determined. Now $f$ is equalized by the two face maps, so $f$ factors through the ...


1

There is a tensor-hom adjunction for the tensor product of algebras, but it exists at the level of the Morita 2-category, rather than the 1-category of algebras. Namely, the Morita 2-category has objects $k$-algebras, and the category of morphisms $A \to B$ is the category $\text{Mod}(A^{op} \otimes B)$ of $(A, B)$-bimodules, where composition is given by ...



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