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5

Day convolution is a categorification of the monoid algebra construction. There is a formal analogy between the two, but one is not a literal generalisation of the other. So to address your question 3, we should not expect to recover the usual convolution from Day convolution. Let's develop the following analogy: \begin{array}{|c|c|} \hline \textbf{monoid ...


1

I don't know if this can be of help, but what I found really useful to gain an intuition behind Day convolution is the correspondence between convolution products and promonoidal structures; the two things can be identified, as every convolution arises from a single promonoidal structure (this dates back to the work of Day himself, and I stated the result in ...


2

Hmm, a logical view of presheaves is as categorified predicates. If we choose the source category as discrete, then we can interpret the coend formula as a categorification of an existential quantification (if our presheaves only return {} or {$*$} then it will be exactly existential quantification.) A discrete monoidal category is a monoid. The coend ...


0

To answer your question, category theory is not ambiguous. It is a viable alternative to set theory and an alternative foundation for all of mathematics. It was first proposed over half a century ago and is a generally accepted framework (unlike other theories such as fuzzy sets for example, which still seem to face an uphill battle). It has been very ...


1

So, as you've noticed, this notion is absolutely ubiquitous. There's a general notion of "isomorphic objects are equal (for all intents and purposes)". Particularly for categorists, notions that distinguish between isomorphic (or more generally, equivalent) objects are often referred to as "evil". In category theory, the principle you describe is called ...


3

I'm sure there will be a better answer, but here are my thoughts. It happens that books on mathematics talk a lot and miss some rigorous treatment. I see two reasons for that. Firstly, usually careful rigorous explanation is useless for the purposes of the theory; neither does it provide insight, nor makes things or intentions clear. Secondly, the reader is ...


3

For a category $C$, and two objects $X,Y$ in $C$, the hom-set (for homomorphism set) is given by $$C(X,Y) = \{f \mid f \text{ is an arrow } f: X \rightarrow Y \text{ in } C\}.$$ Now, despite the name, there is no a priori reason why this should be a set, rather than a class. Indeed, it is not difficult to give an example where a hom-set is a proper class ...


2

The fact that all internal co-categories in a coherent category are necessarily co-equivalence relations [see Peter Lumsdaine's TAC article A small observation on co-categories] provides a telltale sign that the category of posets fails to be a topos. For the inclusion functor $\textbf{Poset} \to \textbf{Cat}$ is represented by the internal co-category ...


2

The category of posets has no subobject classifier. Indeed, if there did exist a universal subobject $1\to \Omega$, then it would be a regular subobject, and so every subobject would be regular since regular subobjects are stable under pullback. But the regular subobjects are exactly the "subposets" in the usual sense (i.e., subsets with the inherited ...


2

Think of a double category as having horizontal arrows as the arrows of $J_0$, vertical arrows as the objects of $J_1$, and squares between them as the arrows of $J_1$. Most of the examples I know are formal, i.e. I don't think double categories are used as much as 2-categories and bicategories outside of category theory proper, but you can find several on ...


1

Let $0 \to X \to Y$ be a sequence in $\mathcal B$ such that $$\operatorname{Hom}(Y,B) \to \operatorname{Hom}(X,B) \to 0$$ is exact. Now let $T$ be the kernel of $X \to Y$. The composition $T \to X \to Y$ is zero, hence the composition $$\operatorname{Hom}(Y,B) \to \operatorname{Hom}(X,B) \to \operatorname{Hom}(T,B)$$ is zero. By the surjectivity of the ...


2

The claim follows from this fact: If $H$ is a faithful contravariant functor and $H f$ is an epimorphism, then $f$ is a monomorphism. As you say, $\mathrm{Hom}(-, B)$ is faithful, so if $\mathrm{Hom} (F A', B) \to \mathrm{Hom} (F A, B)$ is an epimorphism, then $F A \to F A'$ is a monomorphism. But $\mathrm{Hom} (F {-}, B)$ is exact and $A \to A'$ is a ...


3

The statement that a natural transformation is a map between functors is wrong, simple because functors are not sets (and even if they were realised as sets, it is certainly not those sets that the natural transformation maps between). But natural transformations can be thought of as morphisms between functors; thus one can define a category whose objects ...


1

A natural transformation $\eta : F \to G$ between two functors $F,G : \mathbf{C} \to \mathbf{D}$ is an $\operatorname{Ob}(\mathbf{C})$-indexed family $\left(\eta_{c}\right)_{c \in \mathbf{C}}$ of morphisms $\eta_c : F(c) \to G(c)$ (necessarily in $\mathbf{D}$), such that the naturality squares commute. To answer your first question, note that to index a ...


1

In the context of category theory, all "maps" are morphisms. There are no other "maps". If, however, you happen to know that the morphisms in your category correspond to morphisms in another category (typically via a functor), then you can talk about whether a product in one category is a product in the corresponding category. Put a little more formally, ...


1

Not all categories are of the form "objects = sets with structure", and "morphism = structure preserving map/function between such sets". In general category theory this question makes no sense, cannot even be asked. There are only "morphisms" (or arrows) between objects and these objects need not be sets, nor need the arrows/morphisms be functions. Of ...


1

"Does it mean that of all morphisms, only one satisfies this, but there could be a map which is not a morphims which makes this work?" Why are you making a distinction between maps and morphisms? By the definition of a category, there is a set of morphisms $Hom_D(Z,X\times Y)$. The condition that $X\times Y$ is the product of $X$ and $Y$ is just saying that ...


0

You can think of it as a function which indexes morphisms in the category $\mathcal{D}$ with the indexing set being objects in $\mathcal{C}$, i.e. we have that a natural transformation $\eta: Ob(\mathcal{C})\to Arr(\mathcal{D})$ such that $\eta (x)= \eta_x:F(x)\to G(x)$ also written as $\eta(x)\in$ Hom$_{\mathcal{D}}(F(x), G(x))$. Here we can see that the ...


2

Category theory allows us to prove theorems that apply to many different mathematical theories. For example, the categorical theorem that says that left adjoint functors commute with colimits can be applied to prove that the tensor product is right exact and to prove that the suspension functor commutes with wedge products. Those results can be (easily) ...


2

It isn't merely the data structure that makes categories "special" but the operations that can be performed on them. Functors, in particular, relate corresponding elements in different categories and can show how apparently different mathematical structures are—generally at a very abstract level—in fact the same.


1

I see nothing "magical". A topological space is a set with certain additional properties. The category of topological spaces respects all properties of the category of sets.


2

As you mentioned in your previous questions, $\mathbf{Cat}$ is equivalent to $JA-\mathbf{Cat}$. To understand anything categorical about $JA-\mathbf{Cat}$ is to understand it about $\mathbf{Cat}$, because all categorical properties are stable under equivalence. However, the wide subcategory of $\mathbf{Cat}$, whose arrows are (any types of) embeddings is ...


2

Here is a counterexample for finite (co)completeness when $n = 2$. Consider the two cobordisms $\alpha, \beta : S^1 \to S^1$ given respectively by a cylinder and a cylinder with a hole. Then the equalizer of $\alpha$ and $\beta$ does not exist, because there doesn't even exist a cobordism $\gamma : C \to S^1$ (for some compact 1-manifold $C$) such that ...


2

It is easy to see that every idempotent splits in your category: if $e:C\to C$ is idempotent and fully faithful, let $B$ be the full subcategory of $C$ spanned by the objects in the image of $C$. Then the inclusion $B\to C$ is fully faithful, and $e$ factors fully faithfully through this inclusion, giving a splitting of $e$. However, your category is not ...


0

Step 1: Suppose $f: V \rightarrow W$ is a morphism of affine varieties, and let $f^{\#}: A(W) \rightarrow A(V)$ be the induced homomorphism of affine coordinate rings. Let $Y$ be a closed irreducible subset of $V$ corresponding to the prime ideal $P$ of $A(V)$. Then the the zero set of $\left(f^{\#}\right)^{-1} (P)$ in $W$ is precisely the closure of $f(V)$ ...


1

No, such functor doesn't exist in general case. For example, let $A=\{1\}$ (the category of arrows, corresponding to the trivial monoid) and $B=\mathbb{Z}/2\mathbb{Z}$ (the category of arrows, corresponding to the multiplicative monoid of integers modulo $2$, whose arrows are $0$ and $1$). Note, that $EQ_A\subset EQ_B$, but the inclusion mapping ...


3

The forgetful functor $\mathsf{LRS} \to \mathsf{RS}$ has a right adjoint. The right adjoint "$\mathrm{Spec}$" is a rather direct generalization of the spectrum of a commutative ring. You can find the construction in W. D. Gillam's Localization of ringed spaces, for instance. The underlying set of $\mathrm{Spec}(X,\mathcal{O}_X)$ consists of all pairs ...


5

Whenever a non-trivial category is cartesian-closed, the final object $1$ cannot also be an initial object. Otherwise: $$\mathrm{Hom}(A,B)\cong \mathrm{Hom}(1\times A,B) \cong \mathrm{Hom}(1,B^A)\cong\{\cdot\}$$ That is, every home set would have to be a singleton.


5

If $M$ is a non-trivial monoid, the functor $M \times - : \mathsf{Mon} \to \mathsf{Mon}$ doesn't preserve the initial object, and hence it cannot have a right adjoint.


1

Your guess is correct. I think that the easiest way to prove this is to note that for some $c,d$, $E^\bullet$ is quasi-isomorphic to a bounded complex of finitely generated projectives concentrated in degrees $[c,d]$ such that the first differential is not a split monomorphism and the last differential is not a (split) epimorphism, and that then $-d,-c$ are ...


2

Strictly speaking the association of $D_f$ to $R_f$ only defines the sections on a basis; to construct the regular functions on an arbitrary open $U$, we have to take the inverse limit of the $D_f$ contained in $U$. The fact that this is a contravariant functor is pretty straightforward from the universal property of inverse limits. Separatedness is also ...


5

Uff. It's hard to talk about this without introducing all of categorical type theory. The general notion is the notion of a comprehension category which I believe was introduced by Bart Jacobs. See his book or his thesis. However, when I did a search to get a reference for some definitions I found this recent set of notes, "Type Theory through ...


3

Recall that a morphism $f : A \to B$ is a monomorphism if it's left-cancellative. This is equivalent to saying that the map $$(f \circ -) : \operatorname{Hom}(-,A) \to \operatorname{Hom}(-,B)$$ is injective. Of course, whether or not this map is injective has little to do with $f$'s concrete representation as a function of sets, and as D_S's answer shows, ...


3

Short answer: yes what you describe are called monoidal categories. The theory of monoidal categories is quite important because there are lots of application and because they are the basis for enriched category theory. Edit There's also a very interesting difference about monoids and monoidal categories: while the structure of a monoid is specified by the ...


8

Even if in the category you're working in, the objects are sets, the morphisms are functions, and the identity morphism is the literal identity map, then monomorphisms don't have to be injective. (Very cheap) counterexample: let $A, B$ be distinct sets, and let $f:A \rightarrow B$ be any function which is not injective. Let $\mathscr C$ be the category ...


18

"Injection" makes sense in a concrete category, namely a category $C$ equipped with a faithful functor $F : C \to \text{Set}$: a morphism $f$ is an injection if $F(f)$ is an injection (equivalently, a monomorphism). Faithful functors always reflect monomorphisms: if $F(f)$ is a monomorphism, then so is $f$. The proof is straightforward. If $fg = fh$, then ...


0

In general it is not possible to extend a given preorder $(X, \sqsubseteq)$ to a preordered monoid $(X, \sqsubseteq, \cdot, e)$: Consider any preorder $(X, \sqsubseteq)$ such that for all $x \in X$ there is some $y \in X$ with $y \sqsubseteq x$ and $x \not \sqsubseteq y$. [For example $(\mathbb Z, \le)$]. Then there is no $e \in X$ such that $e \sqsubseteq ...


1

I have some partial progress in the case where $W$ is affine. First, assume $V, W$ are both affine varieties. Show the natural map given by $t$, $Hom(V, W) \to Hom(tV, tW)$ is equal to the composition \begin{align*} Hom_{Var}(V, W) \to Hom_{fg \text{ } k-alg}(AW, AV) \to Hom_{Sch/k}(Spec AV, Spec AW) \to Hom_{Sch/k}(tV, tW), \end{align*} where the first ...


2

Pullbacks in $\mathbf{CRing}$ do not necessarily go to pushouts in $\mathbf{Sch}$ or $\mathbf{Set}$. Consider the construction of $\mathbb{P}^1_k$: in $\mathbf{Sch}$ (resp. $\mathbf{Set}$), we have the following pushout square, $$\require{AMScd} \begin{CD} \mathbb{A}^1_k \setminus \{ 0 \} @>>> \mathbb{A}^1_k \\ @VVV @VVV \\ \mathbb{A}^1_k ...


2

It will be true if both maps youre pushing out along are closed embeddings. It will also be true if one of the maps is an infinitesimal thickening (= a closed embedding which induces an isomorphism on the reduced closed subschemes). This type of thing is useful in defromation theory. In general, you should be able to find counterexamples pretty easily.


2

Direct limit is generalisation of the notion of union of a family of sets. Two other examples: Lazard's theorem in commutative algebra asserts that a flat $R$-module is a direct limit of free $R$-modules. Germs of continuous function at a point $a$ of a topological space is defined as the direct limit of the system of pairs $(U,f)$, where $U$ is an open ...


2

Given a prime $p$, the Prüfer $p$-group is the direct limit of the cyclic groups of order a power of $p$. If $\Omega$ is an infinite set, the group of finitary permutations on $\Omega$ is the direct limit of all groups of permutations on the finite subsets of $\Omega$.


1

Yes, $k_K(i)=K$ and $k_K(f)=id(K)$ for every $i\in Ob(I)$ and every $f\in Arr(I)$. It is called a constant functor and usually is denoted by $\Delta_K\colon I\to C$. The functor $\Delta\colon C\to C^I$ (which Musa Al-hassy mentioned in comment), such that $\Delta(c)=\Delta_c$, is called a diagonal functor. The object $K\in Ob(C)$ is a limit of the functor ...


2

The answer to your first question is no: as you've observed, we only need that $\operatorname{Hom}_{C}(G,G)$ is a monoid. You could take it to be the free monoid on one generator, which is just $\mathbb{N}$. (This is sort of a general model of a discrete dynamical system.) I think the confusion in your second question arises from a type error. You're ...


3

A one-object category is not necessarily a group. A one-object category where every morphism is invertible corresponds to a group. This seems to be your misconception as far as I can tell, and it resolves both of your problems. Regarding (1), $C$ doesn't necessarily have an object which is a group. When every morphism is invertible, the set of morphisms ...


4

(1) If $\mathbb{Z} \to \mathbb{Z}_p$ was an epimorphism, this would imply that $\mathbb{Q} \to \mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Q} = \mathbb{Q}_p$ is an epimorphism. But $\mathbb{Q}$ is a field and $\mathbb{Q} \to \mathbb{Q}_p$ is not surjective, so this is a contradiction. (2) I think that $R \to \widehat{R}_I$ is almost never an epimorphism. (3) ...


1

In the general definition, you forgot to mention the condition that $\alpha_{x,x}$ is the canonical isomorphism. Given a scheme $X$ and $R$-modules $M|_x$ for every $R$-point $x : \mathrm{Spec}(R) \to X$ equipped with coherence isomorphisms, choose an open affine covering $(u_i : \mathrm{Spec}(R_i) \to X)$ and consider the $R_i$-modules $ M|_{u_i}$. These ...


1

There is a structure theorem (or existence theorem) for limits (resp. colimits) which says they're built by composing the taking of products and equalizers (resp. coproducts and coequalizers). Since colimits commute with colimits, you can lump all the coproducts on one side and all the coequalizers on another. Doing so one way gives you a quotient of a big ...


2

A morphism from $(X,Y)$ to $(X',Y')$ in $C^{op}\times C$ consists of a morphism $X\to X'$ in $C^{op}$ and a morphism $Y\to Y'$ in $C$, which is the same as a morphism $X'\to X$ in $C$ and a morphism $Y\to Y'$ in $C$. So when they write that $f:X'\to X$, they are thinking of $f$ as a morphism in $C$, rather than $C^{op}$; in $C^{op}$, $f$ is going the ...


1

Choose some $h \in \hom(X,Y)$. Then we have functions $f:X' \to X$, $h:X \to Y$, and $g:Y \to Y'$, so the composition $ghf$ maps $X' \to Y'$. Therefore $ghf \in \hom(X',Y')$. So we have a rule that starts with any morphism $h \in \hom(X,Y)$ and builds from it a morphism $ghf \in \hom(X',Y')$. That means we have a mapping $\hom(X,Y) \to \hom(X',Y')$.



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