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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.


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 ...


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, ...


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 ...


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 $z=x$ ...


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 ...


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.


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 ...


2

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 ...


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

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 ...


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 ...


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 ...


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 ...


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 ...


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.


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 ...


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 ...


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 ...


1

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


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 ...


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.


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

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 ...



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