Mixing
Fibered Categories in Groupoids
$begingroup$
I'm reading an article of Aaron Mazel-Gee about Fibered categories in grupoids and there is an example which I don't understand. Here the full article: https://etale.site/writing/stax-seminar-talk.pdf
...and the excerpt containing the example:
The category $mathcal{C}$ is fixed and we consider a pair $(X_0,X_1)$. So this means for me that $X_i in mathcal{C}$ (eg they are objects in $mathcal{C}$).
On the other hand we consider $U in mathcal{C}$.
From this context I don't see what is here $X_0(U)$ and $X_1(U)$?
category-theory groupoids
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
$endgroup$
add a comment |
$begingroup$
I'm reading an article of Aaron Mazel-Gee about Fibered categories in grupoids and there is an example which I don't understand. Here the full article: https://etale.site/writing/stax-seminar-talk.pdf
...and the excerpt containing the example:
The category $mathcal{C}$ is fixed and we consider a pair $(X_0,X_1)$. So this means for me that $X_i in mathcal{C}$ (eg they are objects in $mathcal{C}$).
On the other hand we consider $U in mathcal{C}$.
From this context I don't see what is here $X_0(U)$ and $X_1(U)$?
category-theory groupoids
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
$endgroup$
add a comment |
$begingroup$
I'm reading an article of Aaron Mazel-Gee about Fibered categories in grupoids and there is an example which I don't understand. Here the full article: https://etale.site/writing/stax-seminar-talk.pdf
...and the excerpt containing the example:
The category $mathcal{C}$ is fixed and we consider a pair $(X_0,X_1)$. So this means for me that $X_i in mathcal{C}$ (eg they are objects in $mathcal{C}$).
On the other hand we consider $U in mathcal{C}$.
From this context I don't see what is here $X_0(U)$ and $X_1(U)$?
category-theory groupoids
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
$endgroup$
I'm reading an article of Aaron Mazel-Gee about Fibered categories in grupoids and there is an example which I don't understand. Here the full article: https://etale.site/writing/stax-seminar-talk.pdf
...and the excerpt containing the example:
The category $mathcal{C}$ is fixed and we consider a pair $(X_0,X_1)$. So this means for me that $X_i in mathcal{C}$ (eg they are objects in $mathcal{C}$).
On the other hand we consider $U in mathcal{C}$.
From this context I don't see what is here $X_0(U)$ and $X_1(U)$?
category-theory groupoids
category-theory groupoids
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
edited Jan 28 at 0:52
KarlPeter
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
asked Jan 28 at 0:05
KarlPeterKarlPeter
5841316
5841316
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are two notions of internal groupoid at play here. One notion of a groupoid internal to a category is simply a functor $mathcal G$ from $mathcal C^{mathbf{op}}$ into groupoids. For instance, any ordinary groupoid $mathcal G=G_1rightrightarrows G_0$ gives rise to a groupoid internal to the category of sets given by the functor $$Smapsto mathbf{Set}(S,G_1)rightrightarrows mathbf{Set}(S,G_0)$$
However, in general an "internal groupoid" in this sense, which really doesn't deserve the name, is a pretty worthless notion. Instead, by an internal groupoid one really means an "internal groupoid" $mathcal G$ in the above sense such that the functors $mathcal{C}^{mathrm{op}}to mathbf{Set}$ given by composing $mathcal G$ with the functors $0,1:mathbf{Gpd}to mathbf{Set}$ taking a (small) groupoid to its set of objects, respectively, morphisms, are both representable. This seems to be the way that the Grothendieck school liked to define such things.
However, Mazel-Gee is beginning with a more elementary notion: a groupoid in a sufficiently reasonable category is defined in a diagrammatically identical way to the definition of an ordinary groupoid. The point missing in your excerpt is that such an internal groupoid is exactly the same thing as an internal groupoid in the above sense: a functor into groupoids with representable components. This identification is probably being made unconsciously here. But the point is that a groupoid $mathcal G$ in $mathcal C$ gives rise to a function $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ by mapping each object in $mathcal C$ into the entire diagram $mathcal G$. Formally, $$hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$$
You can and should check that this really gives a completely ordinary small groupoid for each $c$, which is natural in $c$. The other direction of the equivalence is also not too difficult: once you assume that $mathcal G_0$ and $mathcal G_1$ are representable, the source, target, composition, and identity morphisms become representable by the full faithfulness of the Yoneda embedding, together with the fact that it preserves limits (including pullbacks.)
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
$endgroup$
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
add a comment |
$begingroup$
From Section 4.3.1:
$text{Given a ring $it R$, we define the groupoid $it Q(R)$ of (monic) quadratic expressions and changes of variable by} $
$$text{$it Q(R)$} = left{ob(Q(R)) = text{{$x^2 + bx + c : b,c in R$} $cong R times R$ }, \ text{$Hom_{Q(R)}((b',c'),(b,c)) = text{{r $in R$ : $(x + r)^2 + b'(x + r) + c' = x^2 + bx + c$}}$}right}$$
$text{A ring homomorphism R$to S$ determines a functor $it Q(R) to Q(S)$}.$
$text{So, these constructions assemble into a functor Q : Rings $to$ Groupoids.}$
So essentially $X_0$ and $X_1$ are just functors and $X_0(U)$ is a groupoid.
c.f. page 3 of your source
Edit:
...Applying Spec everywhere, we get a groupoid (pair) (Spec $A$, Spec $Gamma$) in AffSch.
...More explicitly and more generally, any pair of objects $(X_0,X_1)$ in a category $C$ is called a groupoid in $C$ if it has the maps
$$s : X_1 to X_0, t : X_1 to X_0, epsilon : X_0 to X_1, i : X_1 to X_1, m : X_1 times_{s,X_0,t} X_1 to X_1$$
that satisfy the obvious identities coming from the definition of a groupoid.
(This is a generalization of the notion of a $it group space object$
in a category, whose contravariant Yoneda functor lands in Groups.)
So, rather than specifically using the functor $Spec$ (yielding a contravariant equivalence between the categories CRing and AffSch), the author generalized this to a pair of objects $(X_0,X_1)$ in an arbitrary category $C$ that have the structure maps mentioned above. i.e., yes, $X_0$ and $X_1$ are called objects since $X_0,X_1 in ob(C)$.
c.f. page 4
answered Jan 28 at 0:36
Victoria MVictoria M
41617
$endgroup$
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
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@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
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I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
add a comment |
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$begingroup$
There are two notions of internal groupoid at play here. One notion of a groupoid internal to a category is simply a functor $mathcal G$ from $mathcal C^{mathbf{op}}$ into groupoids. For instance, any ordinary groupoid $mathcal G=G_1rightrightarrows G_0$ gives rise to a groupoid internal to the category of sets given by the functor $$Smapsto mathbf{Set}(S,G_1)rightrightarrows mathbf{Set}(S,G_0)$$
However, in general an "internal groupoid" in this sense, which really doesn't deserve the name, is a pretty worthless notion. Instead, by an internal groupoid one really means an "internal groupoid" $mathcal G$ in the above sense such that the functors $mathcal{C}^{mathrm{op}}to mathbf{Set}$ given by composing $mathcal G$ with the functors $0,1:mathbf{Gpd}to mathbf{Set}$ taking a (small) groupoid to its set of objects, respectively, morphisms, are both representable. This seems to be the way that the Grothendieck school liked to define such things.
However, Mazel-Gee is beginning with a more elementary notion: a groupoid in a sufficiently reasonable category is defined in a diagrammatically identical way to the definition of an ordinary groupoid. The point missing in your excerpt is that such an internal groupoid is exactly the same thing as an internal groupoid in the above sense: a functor into groupoids with representable components. This identification is probably being made unconsciously here. But the point is that a groupoid $mathcal G$ in $mathcal C$ gives rise to a function $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ by mapping each object in $mathcal C$ into the entire diagram $mathcal G$. Formally, $$hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$$
You can and should check that this really gives a completely ordinary small groupoid for each $c$, which is natural in $c$. The other direction of the equivalence is also not too difficult: once you assume that $mathcal G_0$ and $mathcal G_1$ are representable, the source, target, composition, and identity morphisms become representable by the full faithfulness of the Yoneda embedding, together with the fact that it preserves limits (including pullbacks.)
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
$endgroup$
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
add a comment |
$begingroup$
There are two notions of internal groupoid at play here. One notion of a groupoid internal to a category is simply a functor $mathcal G$ from $mathcal C^{mathbf{op}}$ into groupoids. For instance, any ordinary groupoid $mathcal G=G_1rightrightarrows G_0$ gives rise to a groupoid internal to the category of sets given by the functor $$Smapsto mathbf{Set}(S,G_1)rightrightarrows mathbf{Set}(S,G_0)$$
However, in general an "internal groupoid" in this sense, which really doesn't deserve the name, is a pretty worthless notion. Instead, by an internal groupoid one really means an "internal groupoid" $mathcal G$ in the above sense such that the functors $mathcal{C}^{mathrm{op}}to mathbf{Set}$ given by composing $mathcal G$ with the functors $0,1:mathbf{Gpd}to mathbf{Set}$ taking a (small) groupoid to its set of objects, respectively, morphisms, are both representable. This seems to be the way that the Grothendieck school liked to define such things.
However, Mazel-Gee is beginning with a more elementary notion: a groupoid in a sufficiently reasonable category is defined in a diagrammatically identical way to the definition of an ordinary groupoid. The point missing in your excerpt is that such an internal groupoid is exactly the same thing as an internal groupoid in the above sense: a functor into groupoids with representable components. This identification is probably being made unconsciously here. But the point is that a groupoid $mathcal G$ in $mathcal C$ gives rise to a function $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ by mapping each object in $mathcal C$ into the entire diagram $mathcal G$. Formally, $$hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$$
You can and should check that this really gives a completely ordinary small groupoid for each $c$, which is natural in $c$. The other direction of the equivalence is also not too difficult: once you assume that $mathcal G_0$ and $mathcal G_1$ are representable, the source, target, composition, and identity morphisms become representable by the full faithfulness of the Yoneda embedding, together with the fact that it preserves limits (including pullbacks.)
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
$endgroup$
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
add a comment |
$begingroup$
There are two notions of internal groupoid at play here. One notion of a groupoid internal to a category is simply a functor $mathcal G$ from $mathcal C^{mathbf{op}}$ into groupoids. For instance, any ordinary groupoid $mathcal G=G_1rightrightarrows G_0$ gives rise to a groupoid internal to the category of sets given by the functor $$Smapsto mathbf{Set}(S,G_1)rightrightarrows mathbf{Set}(S,G_0)$$
However, in general an "internal groupoid" in this sense, which really doesn't deserve the name, is a pretty worthless notion. Instead, by an internal groupoid one really means an "internal groupoid" $mathcal G$ in the above sense such that the functors $mathcal{C}^{mathrm{op}}to mathbf{Set}$ given by composing $mathcal G$ with the functors $0,1:mathbf{Gpd}to mathbf{Set}$ taking a (small) groupoid to its set of objects, respectively, morphisms, are both representable. This seems to be the way that the Grothendieck school liked to define such things.
However, Mazel-Gee is beginning with a more elementary notion: a groupoid in a sufficiently reasonable category is defined in a diagrammatically identical way to the definition of an ordinary groupoid. The point missing in your excerpt is that such an internal groupoid is exactly the same thing as an internal groupoid in the above sense: a functor into groupoids with representable components. This identification is probably being made unconsciously here. But the point is that a groupoid $mathcal G$ in $mathcal C$ gives rise to a function $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ by mapping each object in $mathcal C$ into the entire diagram $mathcal G$. Formally, $$hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$$
You can and should check that this really gives a completely ordinary small groupoid for each $c$, which is natural in $c$. The other direction of the equivalence is also not too difficult: once you assume that $mathcal G_0$ and $mathcal G_1$ are representable, the source, target, composition, and identity morphisms become representable by the full faithfulness of the Yoneda embedding, together with the fact that it preserves limits (including pullbacks.)
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
$endgroup$
There are two notions of internal groupoid at play here. One notion of a groupoid internal to a category is simply a functor $mathcal G$ from $mathcal C^{mathbf{op}}$ into groupoids. For instance, any ordinary groupoid $mathcal G=G_1rightrightarrows G_0$ gives rise to a groupoid internal to the category of sets given by the functor $$Smapsto mathbf{Set}(S,G_1)rightrightarrows mathbf{Set}(S,G_0)$$
However, in general an "internal groupoid" in this sense, which really doesn't deserve the name, is a pretty worthless notion. Instead, by an internal groupoid one really means an "internal groupoid" $mathcal G$ in the above sense such that the functors $mathcal{C}^{mathrm{op}}to mathbf{Set}$ given by composing $mathcal G$ with the functors $0,1:mathbf{Gpd}to mathbf{Set}$ taking a (small) groupoid to its set of objects, respectively, morphisms, are both representable. This seems to be the way that the Grothendieck school liked to define such things.
However, Mazel-Gee is beginning with a more elementary notion: a groupoid in a sufficiently reasonable category is defined in a diagrammatically identical way to the definition of an ordinary groupoid. The point missing in your excerpt is that such an internal groupoid is exactly the same thing as an internal groupoid in the above sense: a functor into groupoids with representable components. This identification is probably being made unconsciously here. But the point is that a groupoid $mathcal G$ in $mathcal C$ gives rise to a function $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ by mapping each object in $mathcal C$ into the entire diagram $mathcal G$. Formally, $$hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$$
You can and should check that this really gives a completely ordinary small groupoid for each $c$, which is natural in $c$. The other direction of the equivalence is also not too difficult: once you assume that $mathcal G_0$ and $mathcal G_1$ are representable, the source, target, composition, and identity morphisms become representable by the full faithfulness of the Yoneda embedding, together with the fact that it preserves limits (including pullbacks.)
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
answered Jan 28 at 2:34
Kevin CarlsonKevin Carlson
33.7k23372
33.7k23372
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
add a comment |
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
sorry for a maybe stupid question but I'm a bit confused about your double arrow (..."coequilizer-notation" :) ...) notation for a groupoid: $mathcal G=G_1rightrightarrows G_0$: Do these two arrows represent the two maps $t$ and $s$ from the excerpt?
$endgroup$
– KarlPeter
Jan 28 at 3:55
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Another point (the most cruical one): If I follow your explanation then a groupoid $mathcal G$ gives rise for the map $hat{mathcal G}:mathcal{C}^{mathrm{op}}to mathbf{Gpd}$ concretely given by $hat{mathcal G}(c)=mathcal C(c,mathcal G_1)rightrightarrows mathcal C(c,mathcal G_0)$. Your notations correspond in following way to those from the excerpt: $c = U in C, C(-, mathcal G_1) = X_1, C(-, mathcal G_0) = X_0$, right? In this case the notation $X_i(U)$ would be make sence except the statement from the article "any pair of objects $(X_0,X_1)$ in a category C ...".
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
Up to now I don't understand why and in which way for $C(-, mathcal G_i) = X_i$ should hold $X_i in C$ as stated in the quote.
$endgroup$
– KarlPeter
Jan 28 at 4:39
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
...Or do you mean this in the sense that the groupoids $mathcal{G}_i$ are $in C$ and you "identify" $X_i$ with $mathcal{G}_i$? Sounds indeed like Yoneda...
$endgroup$
– KarlPeter
Jan 28 at 4:48
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
$begingroup$
"This seems to be the way that the Grothendieck school liked to define such things." Indeed, according to this question this is precisely what Illusie does (for categories rather than groupoids).
$endgroup$
– Arnaud D.
Jan 28 at 11:55
add a comment |
$begingroup$
From Section 4.3.1:
$text{Given a ring $it R$, we define the groupoid $it Q(R)$ of (monic) quadratic expressions and changes of variable by} $
$$text{$it Q(R)$} = left{ob(Q(R)) = text{{$x^2 + bx + c : b,c in R$} $cong R times R$ }, \ text{$Hom_{Q(R)}((b',c'),(b,c)) = text{{r $in R$ : $(x + r)^2 + b'(x + r) + c' = x^2 + bx + c$}}$}right}$$
$text{A ring homomorphism R$to S$ determines a functor $it Q(R) to Q(S)$}.$
$text{So, these constructions assemble into a functor Q : Rings $to$ Groupoids.}$
So essentially $X_0$ and $X_1$ are just functors and $X_0(U)$ is a groupoid.
c.f. page 3 of your source
Edit:
...Applying Spec everywhere, we get a groupoid (pair) (Spec $A$, Spec $Gamma$) in AffSch.
...More explicitly and more generally, any pair of objects $(X_0,X_1)$ in a category $C$ is called a groupoid in $C$ if it has the maps
$$s : X_1 to X_0, t : X_1 to X_0, epsilon : X_0 to X_1, i : X_1 to X_1, m : X_1 times_{s,X_0,t} X_1 to X_1$$
that satisfy the obvious identities coming from the definition of a groupoid.
(This is a generalization of the notion of a $it group space object$
in a category, whose contravariant Yoneda functor lands in Groups.)
So, rather than specifically using the functor $Spec$ (yielding a contravariant equivalence between the categories CRing and AffSch), the author generalized this to a pair of objects $(X_0,X_1)$ in an arbitrary category $C$ that have the structure maps mentioned above. i.e., yes, $X_0$ and $X_1$ are called objects since $X_0,X_1 in ob(C)$.
c.f. page 4
answered Jan 28 at 0:36
Victoria MVictoria M
41617
$endgroup$
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
$begingroup$
@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
$begingroup$
I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
add a comment |
$begingroup$
From Section 4.3.1:
$text{Given a ring $it R$, we define the groupoid $it Q(R)$ of (monic) quadratic expressions and changes of variable by} $
$$text{$it Q(R)$} = left{ob(Q(R)) = text{{$x^2 + bx + c : b,c in R$} $cong R times R$ }, \ text{$Hom_{Q(R)}((b',c'),(b,c)) = text{{r $in R$ : $(x + r)^2 + b'(x + r) + c' = x^2 + bx + c$}}$}right}$$
$text{A ring homomorphism R$to S$ determines a functor $it Q(R) to Q(S)$}.$
$text{So, these constructions assemble into a functor Q : Rings $to$ Groupoids.}$
So essentially $X_0$ and $X_1$ are just functors and $X_0(U)$ is a groupoid.
c.f. page 3 of your source
Edit:
...Applying Spec everywhere, we get a groupoid (pair) (Spec $A$, Spec $Gamma$) in AffSch.
...More explicitly and more generally, any pair of objects $(X_0,X_1)$ in a category $C$ is called a groupoid in $C$ if it has the maps
$$s : X_1 to X_0, t : X_1 to X_0, epsilon : X_0 to X_1, i : X_1 to X_1, m : X_1 times_{s,X_0,t} X_1 to X_1$$
that satisfy the obvious identities coming from the definition of a groupoid.
(This is a generalization of the notion of a $it group space object$
in a category, whose contravariant Yoneda functor lands in Groups.)
So, rather than specifically using the functor $Spec$ (yielding a contravariant equivalence between the categories CRing and AffSch), the author generalized this to a pair of objects $(X_0,X_1)$ in an arbitrary category $C$ that have the structure maps mentioned above. i.e., yes, $X_0$ and $X_1$ are called objects since $X_0,X_1 in ob(C)$.
c.f. page 4
answered Jan 28 at 0:36
Victoria MVictoria M
41617
$endgroup$
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
$begingroup$
@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
$begingroup$
I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
add a comment |
$begingroup$
From Section 4.3.1:
$text{Given a ring $it R$, we define the groupoid $it Q(R)$ of (monic) quadratic expressions and changes of variable by} $
$$text{$it Q(R)$} = left{ob(Q(R)) = text{{$x^2 + bx + c : b,c in R$} $cong R times R$ }, \ text{$Hom_{Q(R)}((b',c'),(b,c)) = text{{r $in R$ : $(x + r)^2 + b'(x + r) + c' = x^2 + bx + c$}}$}right}$$
$text{A ring homomorphism R$to S$ determines a functor $it Q(R) to Q(S)$}.$
$text{So, these constructions assemble into a functor Q : Rings $to$ Groupoids.}$
So essentially $X_0$ and $X_1$ are just functors and $X_0(U)$ is a groupoid.
c.f. page 3 of your source
Edit:
...Applying Spec everywhere, we get a groupoid (pair) (Spec $A$, Spec $Gamma$) in AffSch.
...More explicitly and more generally, any pair of objects $(X_0,X_1)$ in a category $C$ is called a groupoid in $C$ if it has the maps
$$s : X_1 to X_0, t : X_1 to X_0, epsilon : X_0 to X_1, i : X_1 to X_1, m : X_1 times_{s,X_0,t} X_1 to X_1$$
that satisfy the obvious identities coming from the definition of a groupoid.
(This is a generalization of the notion of a $it group space object$
in a category, whose contravariant Yoneda functor lands in Groups.)
So, rather than specifically using the functor $Spec$ (yielding a contravariant equivalence between the categories CRing and AffSch), the author generalized this to a pair of objects $(X_0,X_1)$ in an arbitrary category $C$ that have the structure maps mentioned above. i.e., yes, $X_0$ and $X_1$ are called objects since $X_0,X_1 in ob(C)$.
c.f. page 4
answered Jan 28 at 0:36
Victoria MVictoria M
41617
$endgroup$
From Section 4.3.1:
$text{Given a ring $it R$, we define the groupoid $it Q(R)$ of (monic) quadratic expressions and changes of variable by} $
$$text{$it Q(R)$} = left{ob(Q(R)) = text{{$x^2 + bx + c : b,c in R$} $cong R times R$ }, \ text{$Hom_{Q(R)}((b',c'),(b,c)) = text{{r $in R$ : $(x + r)^2 + b'(x + r) + c' = x^2 + bx + c$}}$}right}$$
$text{A ring homomorphism R$to S$ determines a functor $it Q(R) to Q(S)$}.$
$text{So, these constructions assemble into a functor Q : Rings $to$ Groupoids.}$
So essentially $X_0$ and $X_1$ are just functors and $X_0(U)$ is a groupoid.
c.f. page 3 of your source
Edit:
...Applying Spec everywhere, we get a groupoid (pair) (Spec $A$, Spec $Gamma$) in AffSch.
...More explicitly and more generally, any pair of objects $(X_0,X_1)$ in a category $C$ is called a groupoid in $C$ if it has the maps
$$s : X_1 to X_0, t : X_1 to X_0, epsilon : X_0 to X_1, i : X_1 to X_1, m : X_1 times_{s,X_0,t} X_1 to X_1$$
that satisfy the obvious identities coming from the definition of a groupoid.
(This is a generalization of the notion of a $it group space object$
in a category, whose contravariant Yoneda functor lands in Groups.)
So, rather than specifically using the functor $Spec$ (yielding a contravariant equivalence between the categories CRing and AffSch), the author generalized this to a pair of objects $(X_0,X_1)$ in an arbitrary category $C$ that have the structure maps mentioned above. i.e., yes, $X_0$ and $X_1$ are called objects since $X_0,X_1 in ob(C)$.
c.f. page 4
answered Jan 28 at 0:36
Victoria MVictoria M
41617
edited Jan 28 at 2:48
answered Jan 28 at 0:36
Victoria MVictoria M
41617
answered Jan 28 at 0:36
Victoria MVictoria M
41617
answered Jan 28 at 0:36
Victoria MVictoria M
41617
41617
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
$begingroup$
@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
$begingroup$
I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
add a comment |
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
$begingroup$
@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
$begingroup$
I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
$begingroup$
Hi, thank you for your answer. Yes, in light of this example with $Q$ as functor it seems plausible to transfer it also to this case. The only point that confuses me in the article is the formulation "...any pair of objects $(X_0,X_1)$ in a category $C$ blabla ...". Do you see a reason why the author calls the $X_i$ as "objects" (eg element from $ob(C)$) in $C$?
$endgroup$
– KarlPeter
Jan 28 at 0:49
$begingroup$
@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
$begingroup$
@KarlPeter My comment was too long so I have edited my answer to include my response to your question
$endgroup$
– Victoria M
Jan 28 at 2:40
$begingroup$
I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
I quite don't understand it. If we go back and consider the example with $Spec$ and apply to it this formalism, then we have for $mathcal{C} = $ AffSch and $(X_0,X_1)= (Spec A, Spec Gamma)$. Obviously this pair fullfil the identities with $s, epsilon, i , m$. Futhermore we take an $U in mathcal{C} = $ AffSch. What is then $SpecA(U)$?
$endgroup$
– KarlPeter
Jan 28 at 3:23
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
$begingroup$
It seems that $X_i$ has a kind of double meaning. As an object in $C$ as well as a functor $X_i$ from $C$.
$endgroup$
– KarlPeter
Jan 28 at 3:28
add a comment |
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