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19

"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 ...


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 ...


6

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 ...


5

In a category with all kernel pairs and coequalisers of kernel pairs, the following conditions are equivalent: regular epimorphisms are stable under composition; regular epimorphisms coincide with strong epimorphisms; for any morphism $f$, if $m_f \circ e_f$ is its factorisation through the coequaliser of its kernel pair, $m_f$ is a monomorphism; regular ...


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.


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

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

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 ...


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, ...


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

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 ...


3

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 ...


3

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 ...


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

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 ...


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

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 ...


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

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

There are two different categories: The category $\mathsf{gVec}$ where objects are vector spaces $V$ equipped with a decomposition $V \cong \bigoplus_{n \in \mathbb{Z}} V_n$, and morphisms are linear maps respecting the decomposition; The category $\mathsf{Vec}^\mathbb{Z}$ of families of vector spaces $\{ V_n \}_{n \in \mathbb{Z}}$ and morphisms $f : ...


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 ...


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 ...


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

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 ...


2

Free groups are projective in the category of groups (the exact same argument works as for modules), and for any group $G$, you can take the free group $F$ on the underlying set of $G$ and there is a canonical epimorphism $F\to G$. This works much more generally for pretty much any sort of algebraic structure.


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 ...


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.


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) ...


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 ...


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 ...



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