Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
added to groupoid a section on the description in terms of 2-coskeletal Kan complexes.
Chenchang Zhu kindly writes in to say that she is giving a course on (higher) groupoids and is planning to use relevant nLab pages as course material, and hence panning to edit them further, as need be.
Right now she has added to groupoid the explicit definition.
I have added statement and proof of the relation between equivalence of groupoids and weak homotopy equivalence, in a new section Properties – Equivalences of groupoids.
I have added to groupoid a section “Categories of groupoids” (here) which spells out horizontal composition of homotopies/natural transformations.
Then I used this to spell out the proof that, assuming AC, groupoid representations are euqivalently tuples of group representations (here).
This I also copied over to the entry groupoid representation.
I’ve removed the prime off the lower $F_2$ there. Is that what you meant?
@David_Corfield yes, and also the same diagram is now going $\mathcal{G}_1 \rightarrow \mathcal{G}_2$ but it think it should be $\mathcal{G}_1 \rightarrow \mathcal{G}_4$. I suppose this is just a typo but, since i wasn’t completely sure, i did not correct it myself
giorgio s
@David_Corfield regarding the same lemma (here), i don’t understand the line $(F_2 \cdot \eta \cdot F_1)(x) \;\coloneqq\; F_2(\eta(F_1(x))) \,.$
$\eta$ is said to be a homotopy. Homotopies between groupoids are defined some paragraphs before this lemma, and they can act only on the codomain of the functors they transform, therefore $\eta$ should act on $\mathcal{G}_3$. But writing $\eta(F_1(x))$ means it is acting on $\mathcal{G}_2$
@Marc, i am sorry if i am missing something but your correction (that was also necessary) still doesn’t change the fact that $\eta$ is acting on $F_1(x)\in\mathcal{G}_2$ while i think that $\eta$ can only act on elements of $\mathcal{G}_3$
giorgio s
@giorgio: o.k. let’s dissect the statement of Lemma 2.7 step by step:
(1) $F_2$ and $F^{'}_2$ are functors from $\mathcal{G}_2$ to $\mathcal{G}_3$.
(2) So the natural map $\eta : F^{'}_2 \to F_2$ should assign to every object $y \in \mathcal{G}_2$ a morphism $\eta(y) \in \mathcal{G}_3(F^{'}_2(y),F_2(y))$.
(3) Precomposing with $F_1$ gives for every object $x \in \mathcal{G}_1$ first the object $F(x) \in \mathcal{G}_2$ and then (with $y = F_1(x)$ in (2)) a morphism $\eta(F_1(x)) \in \mathcal{G}_3(F^{'}_2(F_1(x)),F_2(F_1(x)))$.
(4) Applying $F_3$ to $\eta(F_1(x))$ from (3) then gives a morphism $F_3(\eta(F_1(x))) \in \mathcal{G}_4(F_3(F^{'}_2(F_1(x))),F_3(F_2(F_1(x))))$.
So I think indices are the correct ones.
@Marc you are right, thank you for the clarification (i wasn’t thinking of $\eta(y)$ as a morphism)
giorgio s
changed all “delooping groupoid” in the page to “delooping groupoid” and will give this its own little page now.
added pointer to:
A groupoid is a category $C$ with a contravariant endofunctor $(-)^{-1}:C^\op \to C$ which is the identity-on-objects and which satisfies $f \circ f^{-1} = 1_B$ and $f^{-1} \circ f = 1_A$ for all morphisms $f:Hom(A, B)$.
removing query box
+– {: .query} Mike: It’s not clear to me that the notion of “free equivalence relation” doesn’t make sense. Can’t I talk about a left adjoint to the forgetful functor from equivalence relations to, say, directed graphs? Maybe sets-equipped-with-a-binary-relation would be more appropriate, but either one works fine.
Ronnie: Are you sure this forgetful functor equivalence relations to directed graphs has a left adjoint? Suppose the directed graph $\Gamma$ has one vertex $x$ and one loop $u:x \to x$. The free groupoid on $\Gamma$ is the group of integers, which as a groupoid is not an equivalence relation.
Toby: But there is still a free setoid (set equipped with an equivalence relation) on $\Gamma$; it is the point. As a groupoid, it is not the same as the free groupoid on $\Gamma$, although it is the same as the free setoid on the free groupoid on $\Gamma$. If there's an advantage to working with groupoids, perhaps it's that the free groupoid functor preserves distinctions that the free setoid functor forgets? (In this case, a distinction preserved or forgotten is that between $\Gamma$ and the point, which as a graph does not have $u$.) =–
Anonymous
I suspect that Toby Bartels might be the original culprit on the nlab behind the use of “setoid” to mean a “set with an equivalence relation”, with other editors later following him in that step.
1 to 25 of 25