34

The answer is no to all questions except 4.
Negative answers to 1,2 and 3:
It is easy to construct a ring $A$ with an element $a$ satisfying:
(i) $a≠0$,
(ii) $a$ is nilpotent,
(iii) for each $n≥1$ there is $y_n\in A$ such that $a=y_n^n$.
For every morphism $\varphi:A\to C$, the image of $a$ inherits properties (ii) and (iii), hence it is zero if $C$ ...

answered Feb 13 '19 at 14:58

Laurent Moret-Bailly

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One example: Let $X$ be a set with a group $G$ acting on it. Consider the diagram, in the category of sets, with $X$ as its only object, but with all the elements of $G$ (considered as permutations of $X$) as morphisms. The limit of this diagram is the subset of $X$ consisting of the points fixed by the group action, and the colimit is the set of orbits of ...

answered Jan 5 '17 at 18:40

Andreas Blass

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I have asked this question on math.stackexchange last year, and got satisfying answer.
(So this construction did not come from me.)
Let $(X,\mathcal O)$ be a topological space, $\mathcal F(X)$ the partialy ordered set of filters on $X$ with respect to inclusions, considered as a small category in the usual way. Given $x\in X$ and $F\in\mathcal F(X)$ let $\...

18

A $T_1$ colimit $X$ of a sequence of compact spaces $X_n$ is compact iff there is some $n$ such that the map $X_n\to X$ is surjective. This condition is obviously sufficient; suppose that it fails. Passing to a subsequence, we may assume that for each $n$, there is a point $x_n\in X$ that is in the image of $X_n$ but not the image of $X_{n-1}$. Choosing ...

17

### Defining abstract varieties and their morphisms over a finitely generated subfield of the base field

This is treated (in much greater generality) in EGA IV$_3$, Théorème 8.8.2. The existence of $X_0$ and $Y_0$ follows from part (ii), whereas the existence of $f_0$ is part (i).
References.
[EGA IV$_3$] A. Grothendieck, Éléments de géométrie algébrique. IV: Étude locale des schémas et des morphismes de schémas. (Troisième partie). Publ. Math., Inst. Hautes ...

answered Oct 5 '18 at 15:50

R. van Dobben de Bruyn

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The heart of the diagram chase is: All the operations and equations involved in an algebra structure on X have target X. (More generally: the target of each operation/equation could be any limit-preserving functor of the algebra-specified-so-far.)
For each of the kinds of “algebra” you describe, the structure is built up by sequentially adding operations ...

answered Feb 17 at 10:42

Peter LeFanu Lumsdaine

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15

Personally I think that the restricted product description should be avoided. It is best to define $\widehat{\mathbb{Z}}$ to be the inverse limit of the system of all quotients $\mathbb{Z}/n$ (without gratuitously factoring $n$ as a product of primes) and then put $\mathbb{A}=(\mathbb{Q}\otimes\widehat{\mathbb{Z}})\times\mathbb{R}$. We can topologise this ...

answered May 9 '12 at 18:02

Neil Strickland

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Well, yes: the left adjoint of a functor $G: C \to D$ is the initial object in the category whose objects are pairs $(H: D \to C, \eta: 1_D \to G H)$ where $\eta$ is a natural transformation, and whose morphisms $(H, \eta) \to (H', \eta')$ are natural transformations $\theta: H \to H'$ such that
$$\begin{array}{ccc} & 1_D & \\\\ {}^{ \eta} \...

answered Nov 25 '12 at 3:37

13

For $X$ an infinite set, the monad $(-)^X$ (induced from the comonoid structure on $X$ with respect to cartesian product) does not preserve reflexive coequalizers. See page 538 of this paper by Adámek, Koubek, and Velebil. Correspondingly, the forgetful functor for the category of algebras does not preserve reflexive coequalizers (hence also cannot ...

answered Jan 5 '13 at 4:00

13

Let me try to turn my comments into an answer (I think it's also essentially what Vladimir was saying). Suppose you have some diagram $F: K \to \mathcal{C}$. To compute the limit of $F$ is the same as computing the right Kan extension $\epsilon_*F$ along the map $\epsilon: K \to \bullet$. The process you're describing is to compute this Kan extension by ...

13

Binary coproducts do not always exist in $\textrm{Noeth}$.
Assume that the coproduct $C=\overline{\mathbb{Q}} \sqcup \overline{\mathbb{Q}}$ exists in $\textrm{Noeth}$. Letting $A=\overline{\mathbb{Q}} \otimes_{\mathbb{Z}} \overline{\mathbb{Q}}$, we then have a canonical ring map $\varphi : A \to C$. Now the idea is to consider some kind of completion of $A$,...

answered Feb 14 '19 at 20:33

François Brunault

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A sufficient condition for a functor $u:I\to J$ to induce a cofinal functor $Tw(I)\to Tw(J)$ is that $u$ is universally cofinal (i.e. any base change of $u$ is cofinal). Another sufficient condition is that $u$ is universally final. In fact, as may be seen in the proof below, we only need that the base change of $u$ (or of $u^{op}$, respectively) along any ...

11

If you don't group complete, then free $E_{\infty}$-spaces are $1$-truncated.
Consequently, for $k > 0$, the answer is "all $E_{\infty}$-spaces". When $k=0$, you'll
get those which are homotopy equivalent to simplicial commutative monoids.

11

Here are some resources, not necessarily limited to limits and colimits:
Since you specifically ask about software engineering, perhaps you can loko at Steve Easterbrooks's slides "An Introduction to Category Theory for Software Engineers" (which I found by using Google, have you tried?). In particular I draw your attention to slide 18.
Databases and ...

11

I may be showing my ignorance of category theory, but don't think this is true. Work on $l^2$. Let $\mathcal{A}_n$ consist of the operators $A$ satisfying $\langle Ae_i, e_j\rangle = 0$ if $\max(i,j) > n$ (so $\mathcal{A}_n$ is isomorphic to the $n\times n$ matrices). The direct limit of the $\mathcal{A}_n$ is the compact operators on $l^2$. Let $P$ be a ...

11

It depends on what you mean by "the 2-category of monoidal categories" and also what you mean by "complete".
The 2-category of monoidal categories and strict monoidal functors is complete as a Cat-enriched category in the sense of enriched category theory, hence also complete as a bicategory.
The 2-category of monoidal categories and strong monoidal ...

11

If you look at the article on quasitoposes, which are locally cartesian closed and therefore satisfy your exactness condition, you'll find a number of examples that are not locally presentable. For example, the category of pseudotopological spaces, the category of bornological sets, the category of equilogical spaces, and (I think) the category of ...

10

This property holds actually for right derivable categories in the sense of:
MR2729017 Reviewed Cisinski, Denis-Charles Catégories dérivables. (French) [Derivable categories] Bull. Soc. Math. France 138 (2010), no. 3, 317–393.
At least under suitable finiteness assumptions on $I$ and $J$. It also holds for arbitrary small categories $I$ and $J$ when ...

10

### Defining abstract varieties and their morphisms over a finitely generated subfield of the base field

A variety $V$ is a finite union of open affine varieties $V_i$, and because $V$ is separated (usually part of the definition of variety) the intersections $V_i\cap V_j$ are also affine. Now $V$ can be reconstructed from the affine varieties $V_i,V_i\cap V_j$ and the maps of affine varieties $V_i\cap V_j\to V_i$. Obviously, this system is defined over a ...

9

Disclaimer: I am not an expert on model categories.
$\newcommand{\MM}{\mathcal{M}}
\newcommand{\pair}{\mathsf{P}}
\newcommand{\dom}{\operatorname{dom}}
\newcommand{\codom}{\operatorname{codom}}
\newcommand{\colim}{\operatorname*{colim}}
\newcommand{\hocolim}{\operatorname*{hocolim}}
\newcommand{\id}{\mathrm{id}}
\newcommand{\op}{\mathrm{op}}
\newcommand{\Map}...

9

Let $D$ be the derived category of k-vector spaces, and let $C$ be the part consisting of bounded chain complexes of finite-dimensional vector spaces. $D = \mathit{Ind}(C)$.
Let $P$ be a finite partially ordered set whose nerve is a circle, and consider the constant diagrams $\underline{C}$ and $\underline{D}$ shaped like $P$. The limit of $\underline{D}$ ...

9

You can get arbitrarily large cardinalities. For instance, let $X$ be any set and consider the poset $I$ of finite partitions of $X$, ordered by refinement. There is a "tautological" filtered system of finite sets indexed by $I$, and $X$ naturally maps to it. Clearly the map from $X$ to the limit of this system is injective (in fact, the limit can ...

9

This question seems to have a few confusions in it. Here are three thoughts that might be helpful. (I'm guessing you are after a version of statement (2).)
(1) "[T]he natural transformation $\mathcal F(-) \to \mathrm{Hom}_S(-,X)$" is an isomorphism. This is what it means when you say "$X$ represents $\mathcal F$". Isomorphism commute with everything ...

ag.algebraic-geometry reference-request ct.category-theory representable-functors limits-and-colimits

answered Oct 20 '15 at 5:41

Theo Johnson-Freyd

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9

$\Bbb R$ is colimit-dense in the category of real vector spaces but not dense
(see 6.F, 6.34 in my book with J. Adámek "Locally presentable and accessible categories").

9

Yes. Intuitively, this is because all the operations of a monoidal category are "finitary".
One way to prove this formally is that the 2-monad for monoidal categories can be given a presentation in the category of finitary 2-monads, hence it is finitary (which is equivaent to its forgetful functor preserving filtered colimits). A good reference for ...

9

You’re asking whether the functor $M_2$ on Banach spaces preserves colimits of direct sequences.
(In case you’re not familiar with the categorical terminology I’m using here, don’t be put off — it’s not very deep, it’s just useful packaging-up of the kind of properties you’re discussing in the question, and hopefully this answer will give an idea of why ...

answered Oct 12 '18 at 8:27

Peter LeFanu Lumsdaine

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Edit: Alexander Campbell points out in the comments that a reference for this, in the case ${\cal V}=\rm Set$, is Claudio Pisani's paper Sequential multicategories. Following is a sketch of the argument.
Consider the category $\mathcal{V}\text{-}\mathrm{Mult}$ of $\mathcal{V}$-enriched multicategories. This includes the category $\mathcal{V}\text{-}\...

reference-request ct.category-theory monoidal-categories limits-and-colimits enriched-category-theory

8

Such an inverse limit can have any size at all, by choosing the $S_i$ and $I$ properly. Specifically, every set is the inverse limit of a system constructed from its finite subsets. Let $X$ be any nonempty set (the result is clear for the empty set), with element $0\in X$, and let $I$ be the set of finite subsets of $X$ containing $0$, ordered under reverse ...

answered Jun 26 '14 at 13:02

Joel David Hamkins

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There’s a standard property that closely matches what you ask for. For a category $C$ with a functor $U : C \to \newcommand{\Set}{\mathrm{Set}}\Set$ (think of the “forgetful” functor from a category of algebraic structures), we say $U$ creates direct limits over directed sets if (roughly) all such direct limits exist in $C$ and are computed the same way as ...

answered Oct 9 '19 at 8:18

Peter LeFanu Lumsdaine

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The Springer Lecture Notes 639 Topological Vector Spaces of Adasch, Ernst, and Keim contain in § 4 a more or less explicit construction of inductive (=co-) limits in the category of topological vector spaces based on the notion of a string: A sequence $(U_n)_{n\in\mathbb N}$ of balanced and absorbing sets such that $U_{n+1}+U_{n+1}\subseteq U_n$. Similarly ...

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