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13 votes
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Nontrivial advantages of thinking of groups as groupoids with one object?

One statement you can prove by viewing groups as one-object groupoids is that, for any group $G$, the product functor $G\times-\colon\mathrm{Grp}\to\mathrm{Grp}$ preserves connected colimits, so in ...
Daniël Apol's user avatar
  • 5,617
5 votes

Nontrivial advantages of thinking of groups as groupoids with one object?

$\newcommand{\B}{\mathbf{B}}$Another fascinating thing is Mackey's formula. In fact this groupoidal view is the only way I have to this day of remembering the details of the formula at all! As a ...
Trebor's user avatar
  • 4,707
5 votes
Accepted

Commutativity of second square when first square and outer rectangle commute

No. If $C'$ is the terminal object, then the outer rectangle and right square always commute, so it's easy to cook up a counterexample in which the left square fails to commute. For an explicit ...
Alex Kruckman's user avatar
5 votes
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Natural transformation picking out the map from the initial object

Let $\mathscr{C}^\triangleleft:\simeq[0]\star\mathscr{C}$ be the $\infty$-category obtained from $\mathscr{C}$ by adjoining an initial object, and write $-\infty$ for this initial object. It suffices ...
Daniël Apol's user avatar
  • 5,617
4 votes
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Forgetful functor $V: \underline{\mathbf{PSet}} \rightarrow \underline{\mathbf{Set}}$ is not full

By definition, a functor $F\colon\mathcal{C}\to\mathcal{D}$ is full if, for any $x,y\in\mathcal{C}$ the induced map $\mathcal{C}(x,y)\to\mathcal{D}(Fx,Fy), f\mapsto Ff$ is surjective. Your ...
Daniël Apol's user avatar
  • 5,617
4 votes

Can individual topological space be considered as category?

As people in the comments have noted, there are many ways to view a topological space $X$ as a category. One obvious way is to look at its poset of open sets $\mathcal{O} X$, viewed as a category. ...
Chris Grossack's user avatar
3 votes

What is the product in the category of sets with only the injections as maps?

Products of pairs of sets that aren't both singletons don't exist in your category, which I will prove by contradiction. A product of $X,Y$ would have to be a set $X \cdot Y$ equipped with injections $...
Rhys Steele's user avatar
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3 votes
Accepted

Category of Sets vs Category of Types (& Yoneda Lemma)

As you said, the precise definition of "$\mathbf{Type}$" depends on which type theory we're discussing; more importantly, it also depends on whether you take an external or an internal ...
Naïm Favier's user avatar
3 votes

Forgetful functor $V: \underline{\mathbf{PSet}} \rightarrow \underline{\mathbf{Set}}$ is not full

Suppose $(X,x),(Y,y)$ are pointed sets and $f:X\to Y$ is any function such that $f(x)\neq y$, then $f$ is not in the image of the morphism $$ {\rm Hom}_{\underline{\mathbf{PSet}}}((X,x),(Y,y))\to {\rm ...
Tuvasbien's user avatar
  • 9,257
3 votes

Forgetful functor $V: \underline{\mathbf{PSet}} \rightarrow \underline{\mathbf{Set}}$ is not full

$V$ being full means that for every two objects $(X, x_0), (Y, y_0)$ of $\mathrm{PSet}$ the map $\operatorname{Hom}_{\mathrm{PSet}}((X, x_0), (Y, y_0)) \to \operatorname{Hom}_{\mathrm{Set}}(X, Y)$ ...
Ben Steffan's user avatar
  • 3,734
2 votes
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Compatibility of adjunctions for closed monoidal category

Oh, this is a good question. The answer generalises significantly to the enriched setting too (at least in the $\mathscr{V}$ symmetric case; looking at my own proofs I'd suspect a lot goes out of the ...
FShrike's user avatar
  • 41.3k
2 votes
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Stable categories are tensored over spectra

There should be a way to just derive this from the fact that $\mathscr{C}$ is canonically an object in $\mathsf{Mod}_\mathsf{Sp}(\mathsf{Pr}^L)$, as you said. I was thinking along the lines of "...
Daniël Apol's user avatar
  • 5,617
2 votes

What is the product in the category of sets with only the injections as maps?

This category does not have binary products (or a terminal object, i.e., the empty product). To see this, let $1$ be a set with one element, and let $2$ be a set with two elements. Suppose for ...
Alex Kruckman's user avatar
2 votes
Accepted

In the category defined by $\le$ on $\mathbb{Z}$, every morphism defined is a monomorphism and an epimorphism

If $fh=fg$, then there exist an object $x$ such that $g,h\in \mathrm{Hom}(x,a)$. Since $\mathrm{Hom}(x,a)=\{(x,a)\}$, then $g=h=(x,a)$. Thus, $g=h$ so $f$ is a monomorphism. A symmetric argument shows ...
Arturo Magidin's user avatar
1 vote

What is the difference between direct sum $\bigoplus_{i\in I} M_i$ and restricted product $\prod'_{i \in I}M_i$ of abelian group?

There are many types of products you can do. This only applies to products of infinitely many (normal) subgroups, but also not just to abelian groups. There are different names for these depending on ...
David A. Craven's user avatar
1 vote
Accepted

Are comma categories over categories with pullbacks semi-secretly preorders?

As @NaimFavier points out, general slice categories are not preorders. This already fails for any non-preorder category $\mathcal{C}$ with a terminal object $\ast$, by considering the slice $\mathcal{...
Jonas Linssen's user avatar
1 vote
Accepted

The core of an $\infty$-category and pointwise invertible maps $\Delta^1 \times \partial \Delta^n \cup \{1\} \times \Delta^n \to \underline \hom(B,X)$

The answer to Question 1 is yes. Indeed, by the universal property of the pullback, a map $F\colon X\to Y$ factors through $k(Y)\to Y$ iff the composite $X\to Y\to\tau Y$ factors through $k(\tau Y)\to ...
Daniël Apol's user avatar
  • 5,617
1 vote
Accepted

Inverse and Composition of Bisimulations

The definitions of some operations on relations can be found here. That source calls the inverse the "converse" instead, but it's the same thing. Given a relation $R \subseteq A \times B$, ...
S.C.'s user avatar
  • 2,710
1 vote
Accepted

A full embedding functor from $Sob$ to $Frm$?

The covariant functors $\Omega\colon\mathrm{Sob}^\mathrm{op}\to\mathrm{Frm}$ and $\Omega\colon\mathrm{Sob}\to\mathrm{Frm}^\mathrm{op}=\mathrm{Loc}$, where the latter is the category of locales, are ...
Daniël Apol's user avatar
  • 5,617
1 vote

Higher Coherences and Maps from Colimits

I will prove the answer to your question is yes in case we assume that all maps $F(i)\to c$ are equivalences (though not necessarily identity morphisms). I originally failed to see that my solution ...
Daniël Apol's user avatar
  • 5,617
1 vote

On the definition of monomorphisms in arbitrary categories

I think you are correct, in that a formal definition would have to reference something like size. And as was pointed out in the comments, the way you do this depends on the chosen foundations. In ZF ...
Georgii's user avatar
  • 88
1 vote

limit of diagram of limits can be check in the category of sets?

Suppose we already have the following result: Given any sets $X, Y, S$ and maps $u : X \rightarrow S, v: Y \rightarrow S$ between them, and $S$-maps $f, g: X \rightarrow Y$, i.e., $u = v \circ f = v \...
onRiv's user avatar
  • 1,226
1 vote

Is there a connection between the topology of $\mathbb{R}/\mathbb{Q}$ and the logic fundamental to Smooth Infinitesimal Analysis?

You may be confusing $\mathbb R/\mathbb Z$ and $\mathbb R/\mathbb Q$. The former can indeed be viewed as the interval $[0.1]$ with endpoints identified, but the latter is much more complicated. ...
Mikhail Katz's user avatar
  • 43.3k
1 vote

Concrete examples of internal categories (other than small categories)

I would say Hopf-Algeboids, which are categories/groupoids internal to schemes are good examples. One very important example in Algebraic Topology is the one represented by something called $(MU_*,MU_*...
Reihe27's user avatar
  • 141

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