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There are 2 issues at hand here : one is that you missed an important word in the property, and another one is a (non stupid) size issue. First of all, the property says that any functor $C\to D$ with $D$ cocomplete yields a cocontinuous functor $Psh(C)\to D$. This explains why even if $C$ is cocomplete, it's not its own cocompletion : not any functor ...


3

The point is indeed that the swap map $s:S^m\wedge S^n\to S^n\wedge S^m$ induces the natural automorphism of multiplication by $(-1)^{mn}$ on the corepresented functor $\pi_{m+n}$. This is because $(-1)^{mn}$ is the determinant of the map $\mathbb{R}^{m+n}\to \mathbb{R}^{m+n}$ swapping the last $n$ coordinates with the first $m$, and $s$ is the extension of ...


3

The relevant terms here are completeness and cocompleteness. A quick search for these terms will yield the following information and more: A category is complete if it contains all limits over small diagrams (where the index category is small) and it is cocomplete if the colimit of all small diagrams exists. Indeed the category of Groups is a ...


3

Yes, the product of two abelian categories is abelian. The proof is basically trivial: all of the notions used in the definition of an abelian category (zero objects, kernels, cokernels, monics, etc.) can be checked separately on each coordinate when you have a product of categories (so for instance, $(A,B)$ is a zero object iff both $A$ and $B$ are zero ...


3

By Yoneda, we know that if $x,y \in \mathsf{C}$ are objects in some locally small category, then $x \simeq y$ if and only if the functors represented by these objects are naturally isomorphic (that is, either $\mathsf{C}(x,-) \simeq \mathsf{C}(y,-)$ or $\mathsf{C}(-,x) \simeq \mathsf{C}(-,y)$). Let's apply that to prove that $a^1 \simeq a$. By Yoneda, it ...


2

I think you have some typos in your equivalence, and I guess that you wanted to asking for the equivalence: $F\colon \mathsf{Set}\to\mathsf{Set}$ is finitary iff for every $X$ and $x\in FX$ there is a finite $Y$ and $i\colon Y\to X$ with $x\in Fi[FY]$ (By brackets I mean the direct image notation $Fi[FY] \subseteq FX$) Regarding the general definition of ...


2

Riehl (like other sources) state Yoneda using $C(X, -)$ rather than $C(-,X)$ (of course, there's a dual version with $C(-, X)$ - see Exercise 2.2.i) - I'm not sure if this is a typo or if you really want the dual version. Assuming it's a typo: What I think you're concerned about is the following diagram: $$ \begin{array}{ccc} C(X,X) & \stackrel{\...


2

It looks from your edit as though you now know how to use the results from Sheaves on Manifolds to get the result you want. But if you're specifically interested in the fact about bounded complexes of finitely generated modules for a Noetherian ring, then another method is simply to take a projective resolution of your complex and truncate it in ...


2

I'm assuming that what you're after is the definition and sort of the story behind them, rather than any specific theorems about them. Briefest definition: it's a monoid that, as a one-object category, has binary products and exponentials. Somewhat more concretely, it's a monoid with extra operations $\langle-,-\rangle$ and $(-)^*$ representing pairing and ...


2

An elementary definition of $\mathrm{sk}_n X$ is: the smallest simplicial subset of $X$ containing all nondegenerate simplices of dimension at most $n$. The $n$-skeleton is obtained from the $(n-1)$-skeleton by pushing out all the nondegenerate $n$-simplices along the canonical map from their boundary into the $(n-1)$-skeleton, and this pushout process ...


2

Every endofunctor $T$ on $\mathbf{Set}_{< \aleph_1}$ has an initial algebra. To prove this, here's the key set-theoretical lemma: Lemma 1: Let $A_0\subset A_1\subset A_2\subset\dots$ be a sequence of countably infinite sets with union $A$, such that $A_{n+1}\setminus A_n$ is infinite for each $n$. Then the natural map $\operatorname{colim}T(A_n)\to ...


1

In fact, in any pretopos every monomorphism is regular (nLab link). However, I think that an explicit construction for the category of sets will be more insightful in this case. Let $f: X \to Y$ be a monomorphism in $\mathbf{Set}$, so that means that $f$ is injective. Now we define two functions $Y \to \{0,1\}$. The first function $g: Y \to \{0,1\}$ will ...


1

The left adjoint to $\delta$ is the "connected components" functor. Given a category $\mathcal{C}$, we can define an equivalence relation on $\text{ob}(\mathcal{C})$ by taking the symmetric and transitive closure of the relation $xRy$ $\Leftrightarrow$ there is an arrow $X\to Y$. An equivalence class for this relation is called a connected component of $\...


1

A natural transformation is between two functors, just as a functor is between two categories, i.e. they are simply arrows in a suitable category. The word 'comparison' has no special meaning here. Note that in Goldblatt's definition, $F$ and $G$ are given functors, both $\mathscr C\to\mathscr D$, so that they already assign the object $F(a)$ [resp. $G(a)$] ...


1

$\newcommand{\Z}{\mathbb{Z}}$ The idea is mostly correct, though I'd propose some cleanup. Firstly, when defining your morphism $F(\Z) \to \Z$, you do a case analysis on three possibilities, all of which have $(a, b) \in \Z \times \Z$. That means that the functor should be $F(X) = (1+2+3) \times X \times X$. If it were plus, you'd have an element of $1 + 2 + ...


1

Suppose we are given a natural transformation $\mu: \Delta_K \to \hom(X, D-)$, i.e. a family of maps $\mu_i:K\to \hom(X,Di)$ commuting with all the maps $Df\circ - $. If you fix an element $k\in K$, then for every object $i$ of $I$ you have an arrow $\mu_i(k):X\to Di$ in $\mathcal{C}$, and $$Df\circ \mu_i(k)=(Df\circ -)(\mu_i(k))=((Df\circ -)\circ \mu_i)(k)=\...


1

For each finite dimensional vector space pick a basis. Then send each finite dimensional vector space to its dimension, and each linear map to the matrix representing it with respect to the bases chosen for its source and target.


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Construction of functor from category to one of its skeletons. Let $\mathcal C$ be a category. On objects we have the relation of "being isomorphic" which is evidently an equivalence relation. From every equivalence class choose a representative. If $R\subseteq\mathsf{Ob}(\mathcal C)$ denotes this collection of representatives then let $\mathcal S$ be ...


1

Let's take a classic adjoint: $G$ the forgetful functor from $\mathbf{Ab}$ to $\mathbf{Set}$, and $F$ the functor taking a set to the free Abelian group on it. Let $X$ be any nonempty set and $Y=FX$, the free group on $X$. If $\phi$ is the identity map on $FX$, it corresponds to the map $\phi':X\to FX$ taking each $x\in X$ to the corresponding generator of $...


1

Categorical equivalences are the weak equivalences in a model structure. Such a class is essentially never closed under colimits. It is closed under homotopy colimits, and it is an important technical activity to describe classes of ordinary colimits which coincide with the corresponding homotopy colimits. A simple example is that a pushout of two ...


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I wanted to post the text "darx answered the question on MathOverflow: https://mathoverflow.net/a/343557/461" as an answer, but I got the message "Trivial answer converted to comment". Thus I decided to add a copy and paste of darx's answer: No because you can take $A = \mathbf{Z}[x, 1/(x - n); n\geq 0]$ and $B = \mathbf{Z}[x, 1/x, 1/(x - n); n \geq 2]$ and ...


1

Now take a morphism in $[\mathscr A,\mathscr B]^{op}$. It is a natural transformation $\alpha^{op}: G\to F$ (where $F,G:\mathscr A\to\mathscr B$ are functors). The naturality of $\alpha^{op}$ says this: for any arrow $g:A\to A'$ in $\mathscr A$, we have $$\alpha_{A'}^{op}\circ G(g)=F(g)\circ \alpha_A^{op}.$$ This, I think, is the key mistake. $\alpha^{...


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