19

$\require{AMScd}$
A 1974 paper of R. Lagrange, Amalgamation and epimorphisms in $\mathfrak{m}$-complete Boolean algebras (Algebra Universalis 4 (1974), 277–279, DOI link), settled this affirmatively. In the cited paper, Lagrange shows that for any infinite cardinal $\mathfrak{m}$, the category of $\mathfrak{m}$-complete Boolean algebras with $\mathfrak{m}$-...

14

Yes, it is. The reason is:
every object of your presheaf category is a colimit of representables;
so, every object is a filtered colimit of objects which are finite colimits of representables;
so, applying the definition of a compact object, you get a split monomorphism from your compact object $X$ to a finite colimit $T$ of representables. To conclude, ...

6

Here's a pretty direct way to do this. Choose any presentation by generators $x_1,\ldots,x_n$ and relations $r_1,\ldots,r_m$; say $r_i = z_{i,1} \cdots z_{i,k}$ for $z_{i,1},\ldots,z_{i,k} \in \{x_1^{\pm 1}, \ldots, x_n^{\pm 1}\}$. Firstly, we may assume all $r_i$ have length $k = 3$: the new variables
$$x_{i,j} = z_{i,1} \cdots z_{i,j}$$
for $0 \leq j \leq ...

answered 21 hours ago

R. van Dobben de Bruyn

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6

I believe the answer is yes. Assume, by way of contradiction, that some finitely presented group cannot be so expressed. Then we can choose such a group $G$ where for any generating set of the form $x_1,y_1, x_2,y_2,\ldots, x_n,y_n$ the number of non-permissible relations needed to define $G$ (together with some finite number of permissible relations) is ...

5

I think that Aurelien Djament's answer is essentially correct, but I want to nitpick a bit.
If $\mathcal A$ is any locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is any strong generator of finitely-presentable objects, then every finitely-presentable object $X \in \mathcal A$ lies in the closure of $\mathcal C$ under finite ...

5

Are there any places one has to be careful to not allow large categories?
No. For the purposes of forming the 2-category of algebraic/topological/differentiable stacks, or more generally, some kind of presentable stacks over a large category there are no size issues. Naively, the 2-category of stacks on $S$ is carved out from the presheaf category $[S^{op},\...

4

One way to write the universal property of this object $E(a)$ is as follows:
a map $x \to E(a)$ is the same as an isomorphism class of epimorphism $x \times a \twoheadrightarrow k$ in $Set/x$, that is a diagram
$$ x \times a \twoheadrightarrow k \to x$$
whose composite is the first projection.
So it does "feel like" a subobject classifier, but it is not ...

4

You can always reduce an arbitrary "zigzag" as in the first definition, to one of length two, as in the second definition, by applying the following trick:
Whenever you encounter morphisms $E_{j-1}\to E_j\to E_{j+1}$ or $E_{j-1}\leftarrow E_j\leftarrow E_{j+1}$ that go in the same direction, take their composite.
Whenever you encounter morphisms of the form ...

3

You can usually extend two modules from $\mathcal{O}$ by a module which is not semisimple for the Cartan subalgebra (i.e. fails to be a weight module). See Exercise 3.1. in [J. E. Humphreys, Representations of semisimple Lie algebras in the BGG category $\mathcal{O}$. Graduate Studies in Mathematics, 94].

2

I don't think this has been considered. Mainly I've never seen it, but also there are specific feature of this notions that makes it unlikely to be a relevant category theoretic notion independently of your other conditions:
It is not really a universal property, in the sense that it does not characterize what are morphisms to $\Omega$ as not all morphisms ...

2

1. The Leibniz rule follows immediately from the last description
of derivations as morphisms of commutative rings X:R→u(M).
Indeed, u(M) is the square-zero extension of some R-module M'
(in the traditional sense), i.e., u(M)=R⊕M'.
Now a morphism of commutative rings f:R→R⊕M' in the slice category C/R
(not in C, as is claimed in the main post) necessarily ...

2

This is a translation of Dmitri Pavlov's answer into a more intrinsic, more geometric, and more elementary language. In particular, I will show that the étale topos of a positive-dimensional variety is never the topos of a topological space.
If $X$ is a topological space, then the associated topos $E = \mathbf{Sh}(X)$ is generated by subobjects of the final ...

answered May 24 at 2:03

R. van Dobben de Bruyn

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1

There are many references where model categories, and their connection to homology, are described more. See this MO question for a list. For the example of $Ch(R)$, there are several model structures. Those that have quasi-isomorphisms as weak equivalences capture homological algebra (e.g., a morphism $f_*: C_* \to D_*$ is a weak equivalence if the map on ...

1

A reference is Section 7.1 of Fosco Loregian's very nice book on coends, which treats co/lax co/ends. In particular, see Example 7.1.9 for a proof of the formula
$$\mathrm{Nat}_\mathrm{lax}(F,G)=\int_{A\in\mathcal{C}}\mkern-2.05em\square\mkern+1.0em\mathsf{Hom}_{\mathcal{D}}(F(A),G(A)).$$
Another reference for bicategorical coends is Chapter 2 of Alexander ...

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